Question

If the integral $\int \frac{lnx}{x^3}dx=$ $\frac{f\left(x\right)}{4x^2}+C$ , where $f\left(e\right)=-3$ and $C$ is the constant of integration, then the value of $f\left(e^2\right)$ is equal to

• A. $3$
• B. $-4$
• C. $-5$
• D. $5$

Answer 1

Answer: (C)

$\int \underset{I}{\underbrace{lnx}}\underset{II}{\underbrace{\frac{1}{x^3}}}dx=\left(lnx\right)\left(\frac{-1}{2x^2}\right)-\int \frac{1}{x}\left(\frac{-1}{2x^2}\right)dx$
(Using integration by parts)
$=-\frac{lnx}{2x^2}+\frac{1}{2}\left(\frac{x^{-2}}{-2}\right)+C$
$=-\frac{1}{4x^2}\left(2lnx+1\right)+C$
$\Rightarrow f\left(x\right)=-2lnx-1$
$\therefore f\left(e^2\right)=-2ln\left(e^2\right)-1$
$=-4-1=-5$