求∫dx/(x^4-1)不定积分?
∫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。
=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
=(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