文章目录
积分计算
求导 ⇄ 积分 求导 \rightleftarrows 积分求导⇄积分
求导积分
( t ) ′ = 1 \large (t)’=1 (t )′=1 ∫ t d t = 1 2 t 2 + c \large\int tdt=\frac{1}{2}t^2+c ∫t d t =2 1 t 2 +c ( 1 x ) ′ = − 1 x 2 \large(\frac{1}{x})’=-\frac{1}{x^2}(x 1 )′=−x 2 1 ∫ 1 x 2 d x = − 1 x + c \large\int \frac{1}{x^2}dx=-\frac{1}{x}+c ∫x 2 1 d x =−x 1 +c ( l n x ) ′ = 1 x \large(lnx)’=\frac{1}{x}(l n x )′=x 1 ∫ 1 x d x = ∫ d x x = l n ∣ x ∣ + c \large{\int \frac{1}{x}dx=\int \frac{dx}{x}=ln|x|+c}∫x 1 d x =∫x d x =l n ∣x ∣+c ( x ) ′ = 1 2 x \large(\sqrt{x})’=\frac{1}{2\sqrt{x}}(x )′=2 x 1 ∫ 1 x d x = 2 x + c \large\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+c ∫x 1 d x =2 x +c ( a x ) ′ = a x l n a \large(a^x)’=a^xlna (a x )′=a x l na ∫ a x d x = a x l n a + c \large\int a^xdx= \frac{a^x}{lna}+c ∫a x d x =l na a x +c ( s i n x ) ′ = c o s x \large(sinx)’=cosx (s in x )′=cos x ∫ c o s x d x = s i n x + c \large\int cosxdx=sinx+c ∫cos x d x =s in x +c ( t a n x ) ′ = s e c 2 x \large(tanx)’=sec^2x (t an x )′=se c 2 x ∫ s e c 2 x d x = t a n x + c \large\int sec^2xdx=tanx+c ∫se c 2 x d x =t an x +c ( s e c x ) ′ = ( 1 c o s x ) ′ = s e c x t a n x = − s i n x c o s 2 x \large(secx)’=(\frac{1}{cosx})’=secxtanx=-\frac{sinx}{cos^2x}(sec x )′=(cos x 1 )′=sec x t an x =−co s 2 x s in x ∫ s e c x t a n x d x = s e c x + c \large\int secxtanxdx=secx+c ∫sec x t an x d x =sec x +c ( c o s x ) ′ = − s i n x \large(cosx)’=-sinx (cos x )′=−s in x ∫ s i n x d x = − c o s x + c \large\int sinxdx=-cosx+c ∫s in x d x =−cos x +c ( c o t x ) ′ = − c s c 2 x \large(cotx)’=-csc^2x (co t x )′=−cs c 2 x ∫ c s c 2 x d x = − c o t x + c \large\int csc^2xdx=-cotx+c ∫cs c 2 x d x =−co t x +c ( c s c x ) ′ = ( 1 s i n x ) ′ = − c s c c o t x = − c o s x s i n 2 x \large(cscx)’=(\frac{1}{sinx})’=-csccotx=-\frac{cosx}{sin^2x}(csc x )′=(s in x 1 )′=−cscco t x =−s i n 2 x cos x ∫ c s c c o t d x = − c s c x + c \large\int csccotdx=-cscx+c ∫cscco t d x =−csc x +c ( a r c t a n x ) ′ = 1 1 + x 2 \large(arctanx)’=\frac{1}{1+x^2}(a rc t an x )′=1 +x 2 1 ∫ 1 1 + x 2 d x = a r c t a n x + c \large\int \frac{1}{1+x^2}dx=arctanx+c ∫1 +x 2 1 d x =a rc t an x +c ( a r c c o t x ) ′ = − 1 1 + x 2 \large(arccotx)’=-\frac{1}{1+x^2}(a rcco t x )′=−1 +x 2 1 ∫ − 1 1 + x 2 d x = a r c c o t x + c \large\int -\frac{1}{1+x^2}dx=arccotx+c ∫−1 +x 2 1 d x =a rcco t x +c ( a r c s i n x ) ′ = 1 1 − x 2 \large(arcsinx)’=\frac{1}{\sqrt{1-x^2}}(a rcs in x )′=1 −x 2 1 ∫ 1 1 − x 2 d x = a r c s i n x + c \large\int\frac{1}{\sqrt{1-x^2}}dx=arcsinx+c ∫1 −x 2 1 d x =a rcs in x +c ( a r c c o s x ) ′ = − 1 1 − x 2 \large(arccosx)’=-\frac{1}{\sqrt{1-x^2}}(a rccos x )′=−1 −x 2 1 ∫ − 1 1 − x 2 d x = a r c c o s x + c \large\int -\frac{1}{\sqrt{1-x^2}}dx=arccosx+c ∫−1 −x 2 1 d x =a rccos x +c
t a n 2 x + 1 = ( s i n x c o s x ) 2 + 1 = ( s i n 2 x + c o s 2 x c o s 2 x ) = 1 c o s 2 x = s e c 2 x 同理 c o t 2 x + 1 = c s c 2 x \large tan^2x+1=(\frac{sinx}{cosx})^2+1=(\frac{sin^2x+cos^2x}{cos^2x})=\frac{1}{cos^2x}=sec^2x \ \~\ 同理cot^2x+1=csc^2x\~t a n 2 x +1 =(cos x s in x )2 +1 =(co s 2 x s i n 2 x +co s 2 x )=co s 2 x 1 =se c 2 x 同理co t 2 x +1 =cs c 2 x
∫ t a n 2 x d x = ∫ ( s e c 2 x − 1 ) d x = t a n x − x + c \int tan^2x dx = \int (sec^2x-1 )dx=tanx-x+c \~\~∫t a n 2 x d x =∫(se c 2 x −1 )d x =t an x −x +c
不定积分公式【积分 ∫ 记得 + C 】 \large 不定积分公式【积分\int记得+C】\~不定积分公式【积分∫记得+C 】
3. 和 4. 的常数 a 2 在前 a r c 反三角函数 , 5. 的 x 2 在前 l n ( 组合 ) ; 8. 和 9. 为 x 2 在前 x 2 ( 本身 ) + l n ( x + 本身 ) ， 10. 为 x 2 ( 本身 ) 3.和4.的常数a^2在前arc反三角函数,5.的x^2在前ln(组合); ~8.和9.为\sqrt{x^2}在前 \frac{x}{2} (本身)+ln(x+本身)，10.为\frac{x}{2} (本身)3.和4.的常数a 2 在前a rc 反三角函数,5.的x 2 在前l n (组合);8.和9.为x 2 在前2 x (本身)+l n (x +本身)，10.为2 x (本身)
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- ∫ s e c x d x = ∫ 1 c o s x d x = ∫ d x c o s x = l n ∣ s e c x + t a n x ∣ + C 1.\large{\int secxdx=\int \frac{1}{cosx}dx=\int \frac{dx}{cosx}=ln|secx+tanx|+C }\~1.∫sec x d x =∫cos x 1 d x =∫cos x d x =l n ∣sec x +t an x ∣+C
- ∫ c s c x d x = ∫ 1 s i n x d x = ∫ d x s i n x = l n ∣ c s c x − c o t x ∣ + C 2. \large{\int cscxdx =\int \frac{1}{sinx}dx =\int \frac{dx}{sinx}= ln|cscx-cotx|+C }\~2.∫csc x d x =∫s in x 1 d x =∫s in x d x =l n ∣csc x −co t x ∣+C
- ∫ 1 a 2 − x 2 d x = a r c s i n x a + C 3.\large{ \int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin\frac{x}{a}+C }\~3.∫a 2 −x 2 1 d x =a rcs in a x +C
- ∫ 1 a 2 + x 2 d x = 1 a a r c t a n x a + C 4. \large{\int \frac{1}{a^2+x^2}dx = \frac{1}{a}arctan\frac{x}{a}+C}\~4.∫a 2 +x 2 1 d x =a 1 a rc t an a x +C
- ∫ 1 x 2 ± a 2 d x = l n ∣ x + x 2 ± a 2 ∣ + C 5. \large{\int \frac{1}{\sqrt{x^2 \pm a^2}}dx=ln|x+\sqrt{x^2 \pm a^2}| + C}\~5.∫x 2 ±a 2 1 d x =l n ∣x +x 2 ±a 2 ∣+C
- ∫ 1 1 + e x d x = x − l n ( 1 + e x ) + C = − l n ( 1 + e − x ) + C 6. \large{\int \frac{1}{1+e^x}dx=x-ln(1+e^x)+C=-ln(1+e^{-x})+C }\~6.∫1 +e x 1 d x =x −l n (1 +e x )+C =−l n (1 +e −x )+C
- ∫ 1 a 2 − x 2 d x = 1 2 a l n ∣ a + x a − x ∣ + C = − [ ∫ 1 x 2 − a 2 d x ] = − [ 1 2 a l n ∣ x − a x + a ∣ ] + C 7.\large{\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|+C=-[\int \frac{1}{x^2-a^2}dx]=-[~~\frac{1}{2a}ln|\frac{x-a}{x+a}|~~]+C }\~7.∫a 2 −x 2 1 d x =2 a 1 l n ∣a −x a +x ∣+C =−[∫x 2 −a 2 1 d x ]=−[2 a 1 l n ∣x +a x −a ∣]+C
- ∫ x 2 + a 2 = x 2 x 2 + a 2 + a 2 2 l n ( x + x 2 + a 2 ) + C 8. \large{\int \sqrt{x^2+a^2}= \frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2})+C}\~8.∫x 2 +a 2 =2 x x 2 +a 2 +2 a 2 l n (x +x 2 +a 2 )+C
- ∫ x 2 − a 2 = x 2 x 2 − a 2 − a 2 2 l n ( x + x 2 − a 2 ) + C 9. \large{\int \sqrt{x^2-a^2}= \frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2})+C}\~9.∫x 2 −a 2 =2 x x 2 −a 2 −2 a 2 l n (x +x 2 −a 2 )+C
- ∫ a 2 − x 2 = x 2 a 2 − x 2 + a 2 2 a r c s i n x a + C 10. \large{\int \sqrt{a^2-x^2}= \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a}+C} \~\~\~10.∫a 2 −x 2 =2 x a 2 −x 2 +2 a 2 a rcs in a x +C
因为有常数 C , 同一积分可有多种答案 , 可求导验证正确性 : 因为有常数C,同一积分可有多种答案,可求导验证正确性:因为有常数C ,同一积分可有多种答案,可求导验证正确性:
- ( l n ∣ s e c x + t a n x ∣ ) ′ = 1 s e c x + t a n x ( s e c x t a n x + s e c 2 x ) = s e c x ( t a n x + s e c x ) s e c x + t a n x = s e c x 1.\large(ln|secx+tanx|)’=\frac{1}{secx+tanx}(secxtanx+sec^2x)=\frac{secx(tanx+secx)}{secx+tanx}=secx \~1.(l n ∣sec x +t an x ∣)′=sec x +t an x 1 (sec x t an x +se c 2 x )=sec x +t an x sec x (t an x +sec x )=sec x
- ( l n ∣ c s c x − c o t x ∣ ) ′ = 1 c s c x − c o t x ( − c s c x c o t x + c s c 2 x ) = c s c x ( − c o t x + c s c x ) c s c x − c o t x = c s c x 2.\large(ln|cscx-cotx|)’=\frac{1}{cscx-cotx}(-cscxcotx+csc^2x)=\frac{cscx(-cotx+cscx)}{cscx-cotx}=cscx \~2.(l n ∣csc x −co t x ∣)′=csc x −co t x 1 (−csc x co t x +cs c 2 x )=csc x −co t x csc x (−co t x +csc x )=csc x
- ( a r c s i n x a ) ′ = 1 1 − ( a x ) 2 ∗ 1 a = 1 a 2 − x 2 3.\large(arcsin\frac{x}{a})’=\frac{1}{\sqrt{1-(\frac{a}{x})^2}}*\frac{1}{a}=\frac{1}{\sqrt{a^2-x^2}} \~3.(a rcs in a x )′=1 −(x a )2 1 ∗a 1 =a 2 −x 2 1
- ( 1 a a r c t a n x a ) ′ = 1 a ∗ 1 1 + ( x a ) 2 = 1 a 2 + x 2 4.\large(\frac{1}{a}arctan\frac{x}{a})’=\frac{1}{a}*\frac{1}{1+(\frac{x}{a})^2}=\frac{1}{a^2+x^2} \~4.(a 1 a rc t an a x )′=a 1 ∗1 +(a x )2 1 =a 2 +x 2 1
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( l n ∣ x + x 2 ± a 2 ∣ ) ′ = 1 x + x 2 ± a 2 ∗ ( 1 + 1 2 x 2 ± a 2 ∗ 2 x ) = 1 x + x 2 ± a 2 ∗ ( x 2 ± a 2 + x x 2 ± a 2 ) = 1 x 2 ± a 2 5.\large{(ln|x+\sqrt{x^2 \pm a^2}| )’= \frac{1}{x+\sqrt{x^2 \pm a^2}}(1+\frac{1}{2\sqrt{x^2 \pm a^2}}2x)=\frac{1}{x+\sqrt{x^2 \pm a^2}}*(\frac{\sqrt{x^2 \pm a^2 }+x}{\sqrt{x^2 \pm a^2}})=\frac{1}{\sqrt{x^2 \pm a^2}} }\~5.(l n ∣x +x 2 ±a 2 ∣)′=x +x 2 ±a 2 1 ∗(1 +2 x 2 ±a 2 1 ∗2 x )=x +x 2 ±a 2 1 ∗(x 2 ±a 2 x 2 ±a 2 +x )=x 2 ±a 2 1
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( x − l n ( 1 + e x ) + C ) ′ = 1 − 1 1 + e x ∗ e x = 1 + e x − 1 1 + e x = 1 1 + e x 6.\large{(x-ln(1+e^x)+C)’=1-\frac{1}{1+e^x}*e^x=\frac{1+e^x-1}{1+e^x}=\frac{1}{1+e^x} } \~6.(x −l n (1 +e x )+C )′=1 −1 +e x 1 ∗e x =1 +e x 1 +e x −1 =1 +e x 1
∫ 1 1 + e x d x 多种积分解法 : \large{\int \frac{1}{1+e^x}dx}多种积分解法:\~∫1 +e x 1 d x 多种积分解法:
凑微分： ∫ 1 1 + e x d x = ∫ 1 + e x − e x 1 + e x d x = ∫ 1 d x − ∫ 1 1 + e x d ( 1 + e x ) = x − l n ( 1 + e x ) + C 凑微分：\large{\int \frac{1}{1+e^x}dx}=\large{\int \frac{1+e^x-e^x}{1+e^x}dx}=\large{\int 1dx-\int \frac{1}{1+e^x}d(1+e^x)}=x-ln(1+e^x)+C \~凑微分：∫1 +e x 1 d x =∫1 +e x 1 +e x −e x d x =∫1 d x −∫1 +e x 1 d (1 +e x )=x −l n (1 +e x )+C
提公因子： ∫ 1 1 + e x d x = ∫ 1 e x ( e − x + 1 ) d x = ∫ e − x e − x + 1 d x = − ∫ 1 e − x + 1 ( e − x + 1 ) = − l n ( 1 + e − x ) + C = − l n ( e x + 1 e x ) + C = − ( l n ( e x + 1 ) − l n e x ) + C = x − l n ( 1 + e x ) + C 提公因子：\large{\int \frac{1}{1+e^x}dx=\int \frac{1}{e^x(e^{-x}+1)}dx= \int \frac{e^{-x}}{e^{-x}+1}dx}=-\int \frac{1}{e^{-x}+1}(e^{-x}+1)=-ln(1+e^{-x})+C \ =-ln(\frac{e^x+1}{e^{x}})+C=-(ln(e^x+1)-lne^x)+C=x-ln(1+e^x)+C \~提公因子：∫1 +e x 1 d x =∫e x (e −x +1 )1 d x =∫e −x +1 e −x d x =−∫e −x +1 1 (e −x +1 )=−l n (1 +e −x )+C =−l n (e x e x +1 )+C =−(l n (e x +1 )−l n e x )+C =x −l n (1 +e x )+C
换元法： ∫ 1 1 + e x d x = ∫ e x e x ( 1 + e x ) d x = ∫ 1 e x ( 1 + e x ) d e x 令 t = e x ∫ 1 t ( 1 + t ) d t = ∫ 1 t − 1 1 + t d t = l n e x − l n ( 1 + e x ) + C = x − l n ( 1 + e x ) + C 换元法：\large\int \frac{1}{1+e^x}dx=\int \frac{e^x}{e^x(1+e^x)}dx=\int \frac{1 }{e^x(1+e^x)}de^x ~~\frac{令t=e^x}{}~~ \int \frac{1}{t(1+t)}dt = \int \frac{1}{t}-\frac{1}{1+t}dt \ =lne^x-ln(1+e^x)+C=x-ln(1+e^x)+C \~换元法：∫1 +e x 1 d x =∫e x (1 +e x )e x d x =∫e x (1 +e x )1 d e x 令t =e x ∫t (1 +t )1 d t =∫t 1 −1 +t 1 d t =l n e x −l n (1 +e x )+C =x −l n (1 +e x )+C
- ( 1 2 a l n ∣ a + x a − x ∣ ) ′ = 1 2 a ( l n ∣ a + x ∣ − l n ∣ a − x ∣ ) ′ = 1 2 a ( 1 a + x ∗ 1 − 1 a − x ∗ ( − 1 ) ) [ 复合函数求导 ] = 1 2 a ( ( a − x ) + ( a + x ) a 2 − x 2 ) = 1 2 a ( 2 a a 2 − x 2 ) = 1 a 2 − x 2 7.\large(\frac{1}{2a}ln|\frac{a+x}{a-x}|)’=\frac{1}{2a}(ln|a+x|-ln|a-x|)’=\frac{1}{2a}(\frac{1}{a+x}1 -\frac{1}{a-x}(-1)) [\small{复合函数求导}] \ \large=\frac{1}{2a}(\frac{(a-x)+(a+x)}{a^2-x^2})=\frac{1}{2a}(\frac{2a}{a^2-x^2})=\frac{1}{a^2-x^2} \~7.(2 a 1 l n ∣a −x a +x ∣)′=2 a 1 (l n ∣a +x ∣−l n ∣a −x ∣)′=2 a 1 (a +x 1 ∗1 −a −x 1 ∗(−1 ))[复合函数求导]=2 a 1 (a 2 −x 2 (a −x )+(a +x ))=2 a 1 (a 2 −x 2 2 a )=a 2 −x 2 1
∫ 1 x 2 − a 2 d x = ∫ 1 ( x + a ) ( x − a ) d x = ∫ 1 2 a [ 1 x − a − 1 x + a ] d x = 1 2 a l n ∣ a + x a − x ∣ \int \frac{1}{x^2-a^2}dx=\int \frac{1}{(x+a)(x-a)}dx=\int\frac{1}{2a}[ \frac{1}{x-a}-\frac{1}{x+a}]dx=\frac{1}{2a}ln|\frac{a+x}{a-x}|\~∫x 2 −a 2 1 d x =∫(x +a )(x −a )1 d x =∫2 a 1 [x −a 1 −x +a 1 ]d x =2 a 1 l n ∣a −x a +x ∣
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( x 2 x 2 + a 2 + a 2 2 l n ( x + x 2 + a 2 ) ) ′ = 1 2 x 2 + a 2 + x 2 ( 1 2 ( x 2 + a 2 ) − 1 2 ∗ 2 x ) + a 2 2 1 x + x 2 + a 2 ∗ ( 1 + 1 2 ( x 2 + a 2 ) − 1 2 ∗ 2 x ) = 1 2 x 2 + a 2 + x 2 2 x 2 + a 2 + a 2 2 ( 1 x + x 2 + a 2 ) ( x + x 2 + a 2 x 2 + a 2 ) = 1 2 x 2 + a 2 + x 2 2 x 2 + a 2 + a 2 2 x 2 + a 2 = 1 2 x 2 + a 2 + x 2 + a 2 2 x 2 + a 2 = 1 2 x 2 + a 2 + 1 2 x 2 + a 2 = x 2 + a 2 8.\large (\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2}~)~)’ \ =\frac{1}{2} \sqrt{x^2+a^2}+\frac{x}{2} (\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}2x)+\frac{a^2}{2} \frac{1}{x+\sqrt{x^2+a^2}}(1+\frac{1}{2}{(x^2+a^2)}^{-\frac{1}{2}}*2x) \ =\frac{1}{2} \sqrt{x^2+a^2}+\frac{x^2}{2 \sqrt{x^2+a^2}}+\frac{a^2}{2}(\frac{1}{x+\sqrt{x^2+a^2}})(\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}})=\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2}{2 \sqrt{x^2+a^2}}+\frac{a^2}{2 \sqrt{x^2+a^2}} \ =\frac{1}{2}\sqrt{x^2+a^2}+\frac{x^2+a^2}{2 \sqrt{x^2+a^2}}=\frac{1}{2}\sqrt{x^2+a^2}+\frac{1}{2}\sqrt{x^2+a^2}=\sqrt{x^2+a^2} \~8.(2 x x 2 +a 2 +2 a 2 l n (x +x 2 +a 2 ))′=2 1 x 2 +a 2 +2 x (2 1 (x 2 +a 2 )−2 1 ∗2 x )+2 a 2 x +x 2 +a 2 1 ∗(1 +2 1 (x 2 +a 2 )−2 1 ∗2 x )=2 1 x 2 +a 2 +2 x 2 +a 2 x 2 +2 a 2 (x +x 2 +a 2 1 )(x 2 +a 2 x +x 2 +a 2 )=2 1 x 2 +a 2 +2 x 2 +a 2 x 2 +2 x 2 +a 2 a 2 =2 1 x 2 +a 2 +2 x 2 +a 2 x 2 +a 2 =2 1 x 2 +a 2 +2 1 x 2 +a 2 =x 2 +a 2
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( x 2 x 2 − a 2 − a 2 2 l n ( x + x 2 − a 2 ) ) ′ = x 2 − a 2 8.\large (\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2}ln(x+\sqrt{x^2-a^2}~)~)’ =\sqrt{x^2-a^2} \~8.(2 x x 2 −a 2 −2 a 2 l n (x +x 2 −a 2 ))′=x 2 −a 2
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( x 2 a 2 − x 2 + a 2 2 a r c s i n x a ) ′ = 1 2 a 2 − x 2 + x 2 ( 1 2 a 2 − x 2 ∗ ( − 2 x ) ) + a 2 2 ( 1 1 − ( x a ) 2 ∗ ( 1 a ) ) = 1 2 a 2 − x 2 + − x 2 2 a 2 − x 2 + a 2 2 a 2 − x 2 = 1 2 a 2 − x 2 + a 2 − x 2 2 a 2 − x 2 = 1 2 a 2 − x 2 + 1 2 a 2 − x 2 = a 2 − x 2 10.\large(\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}arcsin\frac{x}{a})’=\frac{1}{2} \sqrt{a^2-x^2}+\frac{x}{2} ( \frac{1}{2\sqrt{a^2-x^2}}(-2x))+\frac{a^2}{2}(\frac{1}{\sqrt{1-(\frac{x}{a})^2}}(\frac{1}{a})) \ =\frac{1}{2} \sqrt{a^2-x^2}+\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}} =\frac{1}{2} \sqrt{a^2-x^2}+ \frac{a^2-x^2}{2\sqrt{a^2-x^2}} \ =\frac{1}{2} \sqrt{a^2-x^2}+\frac{1}{2} \sqrt{a^2-x^2}= \sqrt{a^2-x^2} \~\~\~10.(2 x a 2 −x 2 +2 a 2 a rcs in a x )′=2 1 a 2 −x 2 +2 x (2 a 2 −x 2 1 ∗(−2 x ))+2 a 2 (1 −(a x )2 1 ∗(a 1 ))=2 1 a 2 −x 2 +2 a 2 −x 2 −x 2 +2 a 2 −x 2 a 2 =2 1 a 2 −x 2 +2 a 2 −x 2 a 2 −x 2 =2 1 a 2 −x 2 +2 1 a 2 −x 2 =a 2 −x 2
换元法【两类】 \large 换元法【两类】换元法【两类】
第一类换元法：凑微分 ( 复合函数逆过程 ) 第一类换元法：凑微分(复合函数逆过程)第一类换元法：凑微分(复合函数逆过程)
如 ( f ( g ( x ) ) ) ′ = f ′ ( g ( x ) ) ∗ g ′ ( x ) 如(f(g(x)))’=f'(g(x))*g'(x)如(f (g (x )))′=f ′(g (x ))∗g ′(x )
对应积分 ∫ f ′ ( g ( x ) ) ∗ g ′ ( x ) d x = ∫ f ′ ( g ( x ) ) d g ( x ) 对应积分\int f'(g(x))*g'(x)dx=\int f'(g(x))dg(x)对应积分∫f ′(g (x ))∗g ′(x )d x =∫f ′(g (x ))d g (x )
简单换元看出： 令 t = g ( x ) ∫ f ′ ( t ) d t = f ( t ) + c 简单换元看出：\frac{令t=g(x)}{} \int f'(t)dt=f(t)+c 简单换元看出：令t =g (x )∫f ′(t )d t =f (t )+c
又如：出题 ∫ e Δ Δ ′ d x , 做题转换成 ∫ e Δ d Δ 又如： 出题\int e^\Delta \Delta’ dx,做题转换成\int e^\Delta d\Delta 又如：出题∫e ΔΔ′d x ,做题转换成∫e Δd Δ
∫ e x 2 x d x = 1 2 ∫ e x 2 ∗ 2 x d x = 1 2 ∫ e x 2 d x 2 = 1 2 e x 2 + c \large\int e^{x^2}xdx=\frac{1}{2} \int e^{x^2}*2xdx=\frac{1}{2} \int e^{x^2}dx^2=\frac{1}{2} e^{x^2}+c \~∫e x 2 x d x =2 1 ∫e x 2 ∗2 x d x =2 1 ∫e x 2 d x 2 =2 1 e x 2 +c
∫ e x 1 x d x = 2 ∫ e x 1 2 x d x = 2 ∫ e x d x = 2 e x + c \large\int e^{\sqrt{x}} \frac{1}{\sqrt{x}}dx=2 \int e^{\sqrt{x}} \frac{1}{2\sqrt{x}}dx=2 \int e^{\sqrt{x}} d\sqrt{x}=2e^{\sqrt{x}}+c ∫e x x 1 d x =2 ∫e x 2 x 1 d x =2 ∫e x d x =2 e x +c
∫ e a r c t a n x 1 1 + x 2 d x = ∫ e a r c t a n x d ( a r c t a n x ) = e a r c t a n x + c \large\int e^{arctanx} \frac{1}{1+x^2}dx=\int e^{arctanx} d(arctanx)=e^{arctanx} +c \~∫e a rc t an x 1 +x 2 1 d x =∫e a rc t an x d (a rc t an x )=e a rc t an x +c
补： ( l n l n x ) ′ = 1 l n x 1 x = 1 x l n x 补： (lnlnx)’=\frac{1}{lnx} \frac{1}{x}=\frac{1}{xlnx}补：(l n l n x )′=l n x 1 x 1 =x l n x 1
∫ 1 x l n x l n l n x d x = 1 x l n x l n l n x d x = 1 l n l n x d ( l n l n x ) = l n ( l n l n x ) + c \int \frac{1}{xlnxlnlnx}dx=\frac{\frac{1}{xlnx}}{lnlnx}dx=\frac{1}{lnlnx}d(lnlnx)=ln(lnlnx)+c \~∫x l n x l n l n x 1 d x =l n l n x x l n x 1 d x =l n l n x 1 d (l n l n x )=l n (l n l n x )+c
∫ m x + n a x 2 + b x + c 先分母求导 ( a x 2 + b x + c ) ′ = 2 a x + b , 则分子凑出 2 a x + b \int \frac{mx+n}{ax^2+bx+c}先分母求导(ax^2+bx+c)’=2ax+b,则分子凑出2ax+b \~∫a x 2 +b x +c m x +n 先分母求导(a x 2 +b x +c )′=2 a x +b ,则分子凑出2 a x +b
积累变型： ∫ d x a 2 s i n 2 x + b 2 c o s 2 x = ∫ d x c o s 2 x ( a 2 t a n 2 x + b 2 ) = ∫ s e c 2 x ( a 2 t a n 2 x + b 2 ) d x = ∫ 1 ( a 2 t a n 2 x + b 2 ) d ( t a n x ) = 1 a ∫ 1 ( a 2 t a n 2 x + b 2 ) d ( a t a n x ) = 1 a ( 1 b a r c t a n x a t a n x b ) + C [ a , b ≠ 0 ] 积累变型：\ \large \int \frac{dx}{a^2sin^2x+b^2cos^2x}=\int \frac{dx}{cos^2x(a^2tan^2x+b^2)}=\int \frac{sec^2x}{(a^2tan^2x+b^2)}dx=\int \frac{1}{(a^2tan^2x+b^2)}d(tanx) \ =\frac{1}{a}\int \frac{1}{(a^2tan^2x+b^2)}d(atanx)=\frac{1}{a}(\frac{1}{b}arctanx\frac{atanx}{b})+C ~~~[a,b\ne 0]\~积累变型：∫a 2 s i n 2 x +b 2 co s 2 x d x =∫co s 2 x (a 2 t a n 2 x +b 2 )d x =∫(a 2 t a n 2 x +b 2 )se c 2 x d x =∫(a 2 t a n 2 x +b 2 )1 d (t an x )=a 1 ∫(a 2 t a n 2 x +b 2 )1 d (a t an x )=a 1 (b 1 a rc t an x b a t an x )+C [a ,b =0 ]
∫ 1 − l n x ( x − l n x ) 2 d x = ∫ 1 − l n x x 2 ( 1 − l n x x ) 2 d x = − ∫ 1 ( 1 − l n x x ) 2 d ( 1 − l n x x ) = ( 1 − l n x x ) − 1 + C \large \int \frac{1-lnx}{(x-lnx)^2}dx=\large \int \frac{\frac{1-lnx}{x^2}}{(1-\frac{lnx}{x})^2}dx=-\int\frac{1}{(1-\frac{lnx}{x})^2}d(1-\frac{lnx}{x})=(1-\frac{lnx}{x})^{-1}+C ∫(x −l n x )2 1 −l n x d x =∫(1 −x l n x )2 x 2 1 −l n x d x =−∫(1 −x l n x )2 1 d (1 −x l n x )=(1 −x l n x )−1 +C
凑微分复杂看不出时，试试某部分求导 [ 出题逆向 ] 凑微分复杂看不出时，试试某部分求导[出题逆向]凑微分复杂看不出时，试试某部分求导[出题逆向]
∫ e t a n 1 x s e c 2 1 x x 2 d x 尝试 ( t a n 1 x ) ′ = s e c 2 1 x ∗ ( − 1 x 2 ) = − s e c 2 1 x x 2 \int \frac{e^{tan\frac{1}{x}}sec^2\frac{1}{x}}{x^2}dx \ 尝试(tan\frac{1}{x})’=sec^2\frac{1}{x}*(-\frac{1}{x^2})=-\frac{sec^2\frac{1}{x}}{x^2} \~∫x 2 e t an x 1 se c 2 x 1 d x 尝试(t an x 1 )′=se c 2 x 1 ∗(−x 2 1 )=−x 2 se c 2 x 1
原式 = ∫ e t a n 1 x d t a n 1 x = e t a n 1 x + c 原式=\int e^{tan\frac{1}{x}}dtan\frac{1}{x}=e^{tan\frac{1}{x}}+c 原式=∫e t an x 1 d t an x 1 =e t an x 1 +c
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第二类换元法 第二类换元法第二类换元法
三角代换：【注意换回 x 】 三角代换：【注意换回x】三角代换：【注意换回x 】
a 2 − x 2 , 令 x = a sin t 或 a cos t \sqrt{a^2-x^2},令x=a\sin t或a\cos t a 2 −x 2 ,令x =a sin t 或a cos t
a 2 + x 2 , 令 x = a tan t 或 a cot t \sqrt{a^2+x^2},令x=a\tan t或a\cot t a 2 +x 2 ,令x =a tan t 或a cot t
x 2 − a 2 , 令 x = a sec t 或 a csc t \sqrt{x^2-a^2},令x=a\sec t或a\csc t x 2 −a 2 ,令x =a sec t 或a csc t
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三角代换推导公式 : 三角代换推导公式:三角代换推导公式:
根式代换：令 t = 根式 根式代换：令t=\sqrt{根式}根式代换：令t =根式
∫ x 3 x + 1 d x 令 t = x ∫ t 3 t + 1 2 t d t [ t 2 = x , 2 t d t = d x ] = 2 ∫ t 4 t + 1 d t = 2 ∫ [ ( t − 1 ) ( t 2 + 1 ) + 1 t + 1 ] d t − ∫ t t + 1 d t 2 ∫ t 4 − t + t 3 − t 3 + t 2 − t 2 + t − t + 1 − 1 t + 1 d t ( 多项式除法，凑分母倍数 ) = 2 ∫ t 3 ( t + 1 ) − t 2 ( t + 1 ) + t ( t + 1 ) − 2 ( t + 1 ) + 2 t + 1 d t = 2 ∫ t 3 − t 2 + t − 2 + 2 t + 1 d t = 1 2 t 4 − 2 3 t 3 + t 2 − 2 t + 2 l n ∣ t + 1 ∣ + c = 1 2 x 2 − 2 3 x 3 + x − 4 x + 4 l n ( x + 1 ) + c [ l n x , x > 0 ] \int \frac{\sqrt{x^3}}{\sqrt{x}+1}dx \frac{令t=\sqrt{x}}{} \int \frac{t^3}{t+1}2tdt~~~[t^2=x,2tdt=dx] \ =2\int \frac{t^4}{t+1}dt=2\int [(t-1)(t^2+1)+\frac{1}{t+1}]dt- \int \frac{t}{t+1}dt\ \2\int \frac{t^4-t+t^3-t^3+t^2-t^2+t-t+1-1}{t+1}dt ~~~~~(多项式除法，凑分母倍数) \~\ =2\int \frac{t^3(t+1)-t^2(t+1)+t(t+1)-2(t+1)+2}{t+1}dt \ =2\int t^3-t^2+t-2+\frac{2}{t+1}dt=\frac{1}{2}t^4-\frac{2}{3}t^3+t^2-2t+2ln|t+1|+c \ =\frac{1}{2}x^2-\frac{2}{3}\sqrt{x^3}+x-4\sqrt{x}+4ln(\sqrt{x}+1)+c ~~~~~[lnx ,x >0] \~∫x +1 x 3 d x 令t =x ∫t +1 t 3 2 t d t [t 2 =x ,2 t d t =d x ]=2 ∫t +1 t 4 d t =2 ∫[(t −1 )(t 2 +1 )+t +1 1 ]d t −∫t +1 t d t 2 ∫t +1 t 4 −t +t 3 −t 3 +t 2 −t 2 +t −t +1 −1 d t (多项式除法，凑分母倍数)=2 ∫t +1 t 3 (t +1 )−t 2 (t +1 )+t (t +1 )−2 (t +1 )+2 d t =2 ∫t 3 −t 2 +t −2 +t +1 2 d t =2 1 t 4 −3 2 t 3 +t 2 −2 t +2 l n ∣t +1∣+c =2 1 x 2 −3 2 x 3 +x −4 x +4 l n (x +1 )+c [l n x ,x >0 ]
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反三角函数： 反三角函数：反三角函数：
∫ d x x x 2 − 1 令 x = s e c t ∫ s e c t ∗ t a n t s e c t ∗ t a n t d t [ ( s e c t ) ′ = s e c t ∗ t a n t ; d x = s e c t ∗ t a n t d t ] = ∫ 1 d t = t + c [ x = s e c t = 1 c o s t → t = a r c c o s 1 x ] = a r c c o s 1 x + c \int \frac{dx}{x\sqrt{x^2-1}}~\frac{令x=sect}{}~\int \frac{secttant}{secttant}dt~~[(sect)’=secttant;dx=secttantdt] \ =\int 1dt=t+c ~~[x=sect=\frac{1}{cost} \to~t=arccos\frac{1}{x}] \ =arccos\frac{1}{x}+c \~∫x x 2 −1 d x 令x =sec t ∫sec t ∗t an t sec t ∗t an t d t [(sec t )′=sec t ∗t an t ;d x =sec t ∗t an t d t ]=∫1 d t =t +c [x =sec t =cos t 1 →t =a rccos x 1 ]=a rccos x 1 +c
分部积分法： 分部积分法：分部积分法：
( u v ) ′ = u ′ v + u v ′ ( 逆运算推导分部积分 ) ： (uv)’=u’v+uv’ (逆运算推导分部积分)：(uv )′=u ′v +u v ′(逆运算推导分部积分)：
∫ ( u v ) ′ d x = ∫ u ′ v d x + ∫ u v ′ d x u v = ∫ v d u + ∫ u d v ∫ u d v = u v − ∫ v d u \ \int(uv)’dx=\int u’vdx+\int uv’dx \ uv = \int vdu+\int udv \ \int udv =uv-\int vdu \~∫(uv )′d x =∫u ′v d x +∫u v ′d x uv =∫v d u +∫u d v ∫u d v =uv −∫v d u
即： ∫ u d v = u v − ∫ v d u 即：\int udv=uv-\int vdu 即：∫u d v =uv −∫v d u
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[ 表格法：对角相连，正负相间 ] + ( 上导下积 ) [表格法：对角相连，正负相间] +(上导下积)\~[表格法：对角相连，正负相间]+(上导下积)
∫ x s i n x d x \int xsinxdx ∫x s in x d x
求导 : x → 1 → 0 求导:x ~~~~~\to~~~ 1 ~~~\to~~~ 0 求导:x →1 →0
↘ + ↘ − ~~~~~~~~~~~~~~~~~~\searrow +~~~~~~~~~~~~~\searrow -↘+↘−
积分 : s i n x → − c o s x → − s i n x 积分:sinx \to -cosx \to -sinx 积分:s in x →−cos x →−s in x
∫ x s i n x d x = − x c o s x + s i n x + c \int xsinxdx=-x cosx+sinx+c \~∫x s in x d x =−x cos x +s in x +c
∫ x e x d x \int xe^xdx ∫x e x d x
求导 : x → 1 → 0 求导:~x ~~~~\to~~~~~ 1 ~~~\to~~~ 0 求导:x →1 →0
↘ + ↘ − ~~~~~~~~~~~~~~~~~~\searrow +~~~~~~\searrow -↘+↘−
积分 : e x → e x → e x 积分:e^x~~ \to~~ e^x~~\to~~~~e^x 积分:e x →e x →e x
∫ x e x d x = x e x − e x \int xe^xdx=xe^x-e^x ∫x e x d x =x e x −e x
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记住 或 能推导 : ∫ s e c 3 x d x , ∫ c s c 3 x d x , ∫ e a x s i n ( b x ) d x , ∫ e a x c o s ( b x ) d x , ∫ s i n ( l n x ) d x , ∫ c o s ( l n x ) d x [ 令 t = l n x ] ∫ e a x s i n ( b x ) d x = 1 a 2 + b 2 ∣ ( e a x ) ′ ( s i n ( b x ) ) ′ e a x s i n ( b x ) ∣ + C [ 化成行列式 ] = 1 a 2 + b 2 ( a e a x ∗ s i n b x − e a x ∗ b c o s b x ) + C 记住~或~能推导 : \ \int sec^3xdx,\int csc^3xdx,\int e^{ax}sin(bx)dx,\int e^{ax}cos(bx)dx,\int sin(lnx)dx,\int cos(lnx)dx~~[令t=lnx] \~\ \large\int e^{ax}sin(bx)dx= \frac{1}{a^2+b^2} \begin{vmatrix} (e^{ax})’ & (sin(bx))’ \ e^{ax} & sin(bx) \ \end{vmatrix} +C ~~~~~~\small[化成行列式] \~\ \large= \frac{1}{a^2+b^2}(ae^{ax}sinbx-e^{ax}bcosbx)+C \ \~\~记住或能推导:∫se c 3 x d x ,∫cs c 3 x d x ,∫e a x s in (b x )d x ,∫e a x cos (b x )d x ,∫s in (l n x )d x ,∫cos (l n x )d x [令t =l n x ]∫e a x s in (b x )d x =a 2 +b 2 1 (e a x )′e a x (s in (b x ))′s in (b x )+C [化成行列式]=a 2 +b 2 1 (a e a x ∗s inb x −e a x ∗b cos b x )+C
有理函数积分 \large 有理函数积分有理函数积分
例 1 求 ∫ x ( x + 1 ) ( x + 2 ) ( x + 3 ) d x 例1~ 求 \int \frac{x}{(x+1)(x+2)(x+3)}dx 例1 求∫(x +1 )(x +2 )(x +3 )x d x
设原式 = ∫ ( A x + 1 + B x + 2 + C x + 3 ) d x = ∫ x ( x + 1 ) ( x + 2 ) ( x + 3 ) d x 设原式=\int(\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3})dx= \int \frac{x}{(x+1)(x+2)(x+3)}dx 设原式=∫(x +1 A +x +2 B +x +3 C )d x =∫(x +1 )(x +2 )(x +3 )x d x
由待定系数法： A ( x + 2 ) ( x + 3 ) + B ( x + 1 ) ( x + 3 ) + C ( x + 1 ) ( x + 2 ) = x 由待定系数法：A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)=x \~由待定系数法：A (x +2 )(x +3 )+B (x +1 )(x +3 )+C (x +1 )(x +2 )=x
[ 特殊值法 ] 令 x = − 1 , 2 A = − 1 , 即 A = − 1 2 ; 令 x = − 2 , − B = − 2 , 即 B = 2 ; 令 x = − 3 , 2 C = − 3 , 即 C = − 3 2 [特殊值法] \ 令x=-1,2A=-1,即A=-\frac{1}{2}; \ 令x=-2,-B=-2,即B=2; \ 令x=-3,2C=-3,即C=-\frac{3}{2} \ \~[特殊值法]令x =−1 ,2 A =−1 ,即A =−2 1 ;令x =−2 ,−B =−2 ,即B =2 ;令x =−3 ,2 C =−3 ,即C =−2 3
原式 = ∫ ( − 1 2 x + 1 + 2 x + 2 + − 3 2 x + 3 ) d x = − 1 2 l n ∣ x + 1 ∣ + 2 l n ∣ x + 2 ∣ − 3 2 l n ∣ x + 3 ∣ + c 原式=\large \int(\frac{-\frac{1}{2}}{x+1}+\frac{2}{x+2}+\frac{-\frac{3}{2}}{x+3})dx=-\frac{1}{2} ln|x+1|+2ln|x+2|-\frac{3}{2}ln|x+3|+c \~原式=∫(x +1 −2 1 +x +2 2 +x +3 −2 3 )d x =−2 1 l n ∣x +1∣+2 l n ∣x +2∣−2 3 l n ∣x +3∣+c
例 2 : 分母括号内能分解尽量分解，如 例2:分母括号内能分解尽量分解，如例2 :分母括号内能分解尽量分解，如
∫ 1 ( x + 1 ) ( x 2 − 1 ) d x = ∫ 1 ( x + 1 ) 2 ( x − 1 ) d x = ∫ [ A x + 1 + B ( x + 1 ) 2 + C x − 1 ] d x = = ∫ A ( x + 1 ) ( x − 1 ) + B ( x − 1 ) + C ( x + 1 ) 2 ( x + 1 ) 2 ( x − 1 ) d x 待定系数法： A ( x + 1 ) ( x − 1 ) + B ( x − 1 ) + C ( x + 1 ) 2 = 1 [ 特殊值法 ] 令 x = − 1 , − 2 B = 1 , B = − 1 2 ; 令 x = − 1 , 4 C = 1 , C = 1 4 ; 令 x = 0 , − A − B + C = 1 , A = − B + C − 1 = − 1 4 原式 = − 1 4 l n ∣ x + 1 ∣ + 1 2 1 x + 1 + 1 4 l n ∣ x − 1 ∣ + C \int \frac{1}{(x+1)(x^2-1)}dx=\int \frac{1}{(x+1)^2(x-1)}dx \ =\int[\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-1}]dx==\int\frac{A(x+1)(x-1)+B(x-1)+C(x+1)^2}{(x+1)^2(x-1)}dx \ 待定系数法：A(x+1)(x-1)+B(x-1)+C(x+1)^2=1 [特殊值法] \ 令x=-1,-2B=1,B=-\frac{1}{2}; \ 令x=-1,4C=1,C=\frac{1}{4}; \ 令x=0,-A-B+C=1,A=-B+C-1=-\frac{1}{4} 原式=-\frac{1}{4}ln|x+1|+\frac{1}{2}\frac{1}{x+1}+\frac{1}{4}ln|x-1|+C \~∫(x +1 )(x 2 −1 )1 d x =∫(x +1 )2 (x −1 )1 d x =∫[x +1 A +(x +1 )2 B +x −1 C ]d x ==∫(x +1 )2 (x −1 )A (x +1 )(x −1 )+B (x −1 )+C (x +1 )2 d x 待定系数法：A (x +1 )(x −1 )+B (x −1 )+C (x +1 )2 =1 [特殊值法]令x =−1 ,−2 B =1 ,B =−2 1 ;令x =−1 ,4 C =1 ,C =4 1 ;令x =0 ,−A −B +C =1 ,A =−B +C −1 =−4 1 原式=−4 1 l n ∣x +1∣+2 1 x +1 1 +4 1 l n ∣x −1∣+C
最简时分母项中幂大于 1 ，依据有理函数分解原则 ( 能互相抵消成分子必须具备的可能 ) ∫ 1 ( x + 1 ) ( x 2 + 1 ) d x = ∫ A x + 1 + B x + C x 2 + 1 d x [ ← 此题重点处 ] 待定系数法 ( 通分 ) ： A ( x 2 + 1 ) + ( B x + C ) ( x + 1 ) = 1 A x 2 + A + B x 2 + B x + C x + C = 1 ( A + B ) x 2 + ( B + C ) x + A + C = 1 则 A + B = 0 , B + C = 0 , A + C = 1 即 C = 1 2 , B = − 1 2 ， A = 1 2 原式 = ∫ 1 2 x + 1 + − 1 2 x + 1 2 x 2 + 1 d x = 1 2 ∫ 1 x + 1 + − x + 1 x 2 + 1 d x = 1 2 ∫ 1 x + 1 + 1 x 2 + 1 d x − 1 2 ∗ 1 2 ∫ 1 x 2 + 1 d ( x 2 + 1 ) = 1 2 l n ∣ 1 + x ∣ + 1 2 a r c t a n x + − 1 4 l n ( x 2 + 1 ) + c 最简时分母项中幂大于1，依据有理函数分解原则(能互相抵消成分子必须具备的可能) \~ \large \int \frac{1}{(x+1)(x^2+1)}dx=\int \frac{A}{x+1}+\frac{Bx+C} {x^2+1}dx \small[\leftarrow 此题重点处]\~\ \small待定系数法(通分)：\ A(x^2+1)+(Bx+C)(x+1)=1 \ Ax^2+A+Bx^2+Bx+Cx+C=1 \ (A+B)x^2+(B+C)x+A+C=1 \ 则A+B=0,B+C=0,A+C=1 即C=\frac{1}{2},B=-\frac{1}{2}，A=\frac{1}{2} \ 原式=\large\int \frac{\frac{1}{2}}{x+1}+\frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+1}dx=\frac{1}{2}\int \frac{1}{x+1}+\frac{-x+1}{x^2+1}dx \ =\frac{1}{2}\int \frac{1}{x+1}+\frac{1}{x^2+1}dx-\frac{1}{2}*\frac{1}{2}\int\frac{1}{x^2+1}d(x^2+1) \ =\frac{1}{2}ln|1+x|+\frac{1}{2}arctanx+ -\frac{1}{4}ln(x^2+1)+c \~最简时分母项中幂大于1 ，依据有理函数分解原则(能互相抵消成分子必须具备的可能)∫(x +1 )(x 2 +1 )1 d x =∫x +1 A +x 2 +1 B x +C d x [←此题重点处]待定系数法(通分)：A (x 2 +1 )+(B x +C )(x +1 )=1 A x 2 +A +B x 2 +B x +C x +C =1 (A +B )x 2 +(B +C )x +A +C =1 则A +B =0 ,B +C =0 ,A +C =1 即C =2 1 ,B =−2 1 ，A =2 1 原式=∫x +1 2 1 +x 2 +1 −2 1 x +2 1 d x =2 1 ∫x +1 1 +x 2 +1 −x +1 d x =2 1 ∫x +1 1 +x 2 +1 1 d x −2 1 ∗2 1 ∫x 2 +1 1 d (x 2 +1 )=2 1 l n ∣1 +x ∣+2 1 a rc t an x +−4 1 l n (x 2 +1 )+c
三角函数积分 \large 三角函数积分 \~三角函数积分
∫ s i n 2 x d x 、 ∫ c o s 2 x d x 偶数次幂，降幂 \int sin^2xdx、\int cos^2xdx偶数次幂，降幂∫s i n 2 x d x 、∫co s 2 x d x 偶数次幂，降幂
∫ s i n 2 x d x = 1 2 ∫ 1 − c o s 2 x 2 d 2 x = 1 2 x − 1 4 sin 2 x + c [ 求导检查正确性 ] \int sin^2xdx=\frac{1}{2}\int \frac{1-cos2x}{2}d2x=\frac{1}{2}x-\frac{1}{4}\sin2x+c ~~[求导检查正确性]~~\~∫s i n 2 x d x =2 1 ∫2 1 −cos 2 x d 2 x =2 1 x −4 1 sin 2 x +c [求导检查正确性]
∫ s i n 3 x d x 、 ∫ c o s 3 x d x 奇数次幂凑微分 \int sin^3xdx、\int cos^3xdx奇数次幂凑微分∫s i n 3 x d x 、∫co s 3 x d x 奇数次幂凑微分
∫ s i n 3 x d x = − ∫ s i n 2 x d c o s x = − ∫ ( 1 − c o s 2 x ) d c o s x = ∫ ( c o s 2 x − 1 ) d c o s x = 1 3 c o s 3 x − c o s x + c \int sin^3xdx=-\int sin^2xdcosx=-\int (1-cos^2x)dcosx=\int (cos^2x-1)dcosx=\frac{1}{3}cos^3x-cosx+c \~∫s i n 3 x d x =−∫s i n 2 x d cos x =−∫(1 −co s 2 x )d cos x =∫(co s 2 x −1 )d cos x =3 1 co s 3 x −cos x +c
万能替换公式： 万能替换公式：万能替换公式：
令 u = t a n x 2 → x = 2 a r c t a n u , d x = 2 d u 1 + u 2 令\large u=tan\frac{x}{2} \to x=2arctanu ,~~dx=\frac{2du}{1+u^2} \~令u =t an 2 x →x =2 a rc t an u ,d x =1 +u 2 2 d u
s i n x = 2 t a n x 2 1 + t a n 2 x 2 = 2 u 1 + u 2 c o s x = 1 − t a n 2 x 2 1 + t a n 2 x 2 = 1 − u 2 1 + u 2 sinx=\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{2u}{1+u^2} \~\ cosx=\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{1-u^2}{1+u^2}\~\~s in x =1 +t a n 2 2 x 2 t an 2 x =1 +u 2 2 u cos x =1 +t a n 2 2 x 1 −t a n 2 2 x =1 +u 2 1 −u 2
∫ 1 1 + s i n x + c o s x d x = ∫ 2 d u 1 + u 2 1 + 2 u 1 + u 2 + 1 − u 2 1 + u 2 = 2 2 + 2 u d u = 1 1 + u d u = l n ∣ 1 + u ∣ + c = l n ∣ 1 + t a n x 2 ∣ + c \large\int\frac{1}{1+sinx+cosx}dx=\int \frac{\frac{2du}{1+u^2}}{1+\frac{2u}{1+u^2}+\frac{1-u^2}{1+u^2}}=\frac{2}{2+2u}du=\frac{1}{1+u}du=ln|1+u|+c=ln|1+tan\frac{x}{2}|+c ∫1 +s in x +cos x 1 d x =∫1 +1 +u 2 2 u +1 +u 2 1 −u 2 1 +u 2 2 d u =2 +2 u 2 d u =1 +u 1 d u =l n ∣1 +u ∣+c =l n ∣1 +t an 2 x ∣+c
不定积分 + C , 定积分没有常数 C \large \red{不定积分+C,定积分没有常数C}不定积分+C ,定积分没有常数C
Original: https://blog.csdn.net/m0_58466526/article/details/128754287
Author: violet~evergarden
Title: 高等数学【合集2】
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