Question

求∫dx/(x^4-1)不定积分？

Answer 1

∫dx/(x^4-1)不定积分是-(1/2)arctanx + (1/4)ln|bai(x-1)/(x+1)| + C

解：

^|∫dx/(x^4-1)

=(1/2)∫[1/(x^2-1) -1/(x^2 +1) ] dx

=-(1/2)arctanx + (1/2)∫dx/(x^2-1)

=-(1/2)arctanx + (1/4)∫[ 1/(x-1) -1/(x+1) ] dx

=-(1/2)arctanx + (1/4)ln|(x-1)/(x+1)| + C

所以∫dx/(x^4-1)不定积分是-(1/2)arctanx + (1/4)ln|bai(x-1)/(x+1)| + C。

扩展资料：

1、常用几种积分公式：

(1）∫0dx=c

(2）∫x^udx=(x^(u+1))/(u+1)+c

(3）∫e^xdx=e^x+c

(4）∫sinxdx=-cosx+c

(5）∫a^xdx=(a^x)/lna+c

(6）∫1/xdx=ln|x|+c

2、一般定理

定理1：设f(x)在区间[a,b]上连续，袜森雹那么f(x)在[a,b]上可积。

定理2：设f(x)在区间[a,b]上告帆单调，那么f(x)在[a,b]上可积。

定理3：设f(x)区间[a,b]上有界，且只有有限个间断点，那么f(x)在[a,b]上春弯可积。

Answer 2

该题目中的不拦搏定积分可以雹衡亩将分源森子进行因式分解后求解。

Answer 3

原式＝∫dx/（x²+1）（x²-1）
＝1/2［∫dx/（x²-1）-∫dx（拦宴拿x²+1）］
＝1/4［∫dx/（x-1）-∫dx（x+1）］-tanx/2
＝(1/4)ln｜（x-1）简搭祥唤/(x+1)｜-(1/2)tanx+C

Answer 4

∫1/(x^4-1)dx=∫1/[(x?-1)(x?+1)]dx=(1/2)∫[1/(x?-1)]-[1/(x?+1)]dx=(1/2)∫[1/(x?-1)]dx-(1/2)∫[1/态团(x?+1)]dx=(1/4)∫[1/没册(x-1)]-[1/(x+1)]dx-(1/2)arctanx=(1/枯闭宏4)∫[1/(x-1)]dx-(1/4)∫[1/(x+1)]dx-(1/2)arctanx=(1/4)ln|x-1|-(1/4)ln|x+1|-(1/2)arctanx=(1/4)ln(|x-1|/|x+1|)-(1/2)arctanx

Answer 5

∫dx/(x^4-1)
=(1/2)∫[1/(x^2-1) -1/(x^2 +1) ] dx
=-(1/2)arctanx + (1/2)∫帆喊誉dx/(x^2-1)
=-(1/2)arctanx + (1/4)∫[ 1/(x-1) -1/渗誉(x+1) ] dx
=-(1/2)arctanx + (1/态段4)ln|(x-1)/(x+1)| + C