Question

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

• A. $\ln (2x+\frac{1}{x^2})+c$
• B. $\ln (2x^3+1)+c$
• C. ${\tan }^{-1}\left(\frac{1}{x^{\frac{3}{2}}+1}\right)+\ln \left(\frac{x-1}{x+1}\right)+c$
• D. $\ln (x+\sqrt{\frac{x-1}{x+1}})+c$

Answer 1

The correct option is A $\ln (2x+\frac{1}{x^2})+c$

$\int \frac{2(x^3-1)}{x^3(2x+\frac{1}{x^2})}dx=\int \frac{2(1-x^{-3})dx}{(2x+\frac{1}{x^2})}$

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

$\Rightarrow (2-\frac{2}{x^3})dx=dt$

$\therefore \int \frac{dt}{t}=\ln |t|+c$

$=\ln (2x+\frac{1}{x^2})+c$