Question

If the integral $\displaystyle \int \frac{ln x}{x^{3}}dx=$ $\frac{f \left(x\right)}{4 x^{2}}+C$ , 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)

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