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)
Correct Option: 1
Solution:
$\int \frac{2 x^{3}-1}{x^{4}+x} d x$
$\int \frac{2 x-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x}} d x$
$x^{2}+\frac{1}{x}=t$
$\left(2 x-\frac{1}{x^{2}}\right) d x=d t$
$\int \frac{\mathrm{dt}}{\mathrm{t}}=\ell \mathrm{n}(\mathrm{t})+\mathrm{C}$
$=\ell n\left(x^{2}+\frac{1}{x}\right)+C$