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)

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

Correct Option: 1

Solution:

$\int \frac{2 x^{3}-1}{x^{4}+x} d x$

$\int \frac{2 x-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x}} d x$

$x^{2}+\frac{1}{x}=t$

$\left(2 x-\frac{1}{x^{2}}\right) d x=d t$

$\int \frac{\mathrm{dt}}{\mathrm{t}}=\ell \mathrm{n}(\mathrm{t})+\mathrm{C}$

$=\ell n\left(x^{2}+\frac{1}{x}\right)+C$