The integral

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Question:
The integral $\int \frac{2 x^{3}-1}{x^{4}+x} d x$ is equal to :

(Here $\mathrm{C}$ is a constant of integration)

(1) $\frac{1}{2} \log _{e} \frac{\left|x^{3}+1\right|}{x^{2}}+C$
(2) $\frac{1}{2} \log _{e} \frac{\left(x^{3}+1\right)^{2}}{\left|x^{3}\right|}+C$
(3) $\log _{e}\left|\frac{x^{3}+1}{x}\right|+C$
(4) $\log _{e} \frac{\left|x^{3}+1\right|}{x^{2}}+C$

Correct Option: 3,

Solution:

Given integral, $I=\int \frac{\left(2 x^{3}-1\right) d x}{x^{4}+x}=\int \frac{\left(2 x-x^{-2}\right) d x}{x^{2}+x^{-1}}$

Put $x^{2}+x^{-1}=u \Rightarrow\left(2 x-x^{-2}\right) d x=d u$

$\Rightarrow I=\int \frac{d u}{u}=\log |u|+c=\log \left|x^{2}+x^{-1}\right|+c$

$=\log \left|\frac{x^{3}+1}{x}\right|+c$