Question

$\int \frac{2x^3-1}{x^4+x}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{2}{\log }_e\left|\frac{x^3+1}{x^2}\right|+C$
• B. $\frac{1}{2}{\log }_e\left|\frac{(x^3+1{)}^2}{x^3}\right|+C$
• C. ${\log }_e\left|\frac{x^3+1}{x}\right|+C$
• D. ${\log }_e\left|\frac{x^3+1}{x^2}\right|+C$

Answer 1

The correct option is C ${\log }_e\left|\frac{x^3+1}{x}\right|+C$

Let $I=\int \frac{(2x^3-1)dx}{x^4+x}$
$\Rightarrow I=\int \frac{(2x-x^{-2})dx}{x^2+x^{-1}}$
Put $x^2+x^{-1}=u$
$\Rightarrow (2x-x^{-2})dx=du$
$\therefore I=\int \frac{du}{u}$
$\Rightarrow I={\log }_e|u|+C$
$\Rightarrow I={\log }_e|x^2+x^{-1}|+C$
$\therefore I={\log }_e\left|\frac{x^3+1}{x}\right|+C$