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，那么f(x)在[a,b]上可积。

定理2：设f(x)在区间[a,b]上单调，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]上可积。

Answer 2

该题目中的不定积分可以将分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（x²+1）］
＝1/4［∫dx/（x-1）-∫dx（x+1）］-tanx/2
＝(1/4)ln｜（x-1）/(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/(x?+1)]dx=(1/4)∫[1/(x-1)]-[1/(x+1)]dx-(1/2)arctanx=(1/4)∫[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)∫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