Mathematics

$$\int { \frac { x }{ { x }^{ 4 }-1 }  } dx$$


SOLUTION
$$\displaystyle \int{\dfrac{x}{x^4-1}}dx$$
$$\Rightarrow u=x^2\Rightarrow du=2xdx$$
$$\displaystyle \int{\dfrac{1 du}{2(u^2-1)}}=\dfrac{1}{4}\displaystyle \int{\dfrac{(u+1)-(u-1)}{(u-1)(u+1)}}$$
$$=\dfrac{1}{4}\displaystyle \int{\dfrac{1}{u-1}-\dfrac{1}{u+1}}$$
$$=\dfrac{1}{4}\ln \left(\dfrac{u-1}{u+1}\right)=\dfrac{1}{4}\left(\dfrac{x^2-1}{x^2+1}\right)$$
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Subjective Medium Published on 17th 09, 2020
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