Question

Let $\displaystyle \int \frac{1 - ln x}{x^{2}}dx=f\left(x\right),$ for all positive $x$ . If $f\left(e\right)=\frac{1}{e},$ then $f\left(2\right)+f\left(4\right)$ is equal to

• A. $ln 4$
• B. $ln 2$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$f\left(x\right)=\displaystyle \int \frac{1}{x^{2}}dx-\displaystyle \int ln x\cdot \frac{1}{x^{2}}dx$
$=-\frac{1}{x}-\left\{ln x \left(- \frac{1}{x}\right) + \displaystyle \int \frac{1}{x^{2}} d x\right\}$
$=-\frac{1}{x}+\frac{ln x}{x}+\frac{1}{x}+c=\frac{ln ⁡ x}{x}+c$
Now, $f\left(e\right)=\frac{1}{e}\Rightarrow c=0$
Hence, $f\left(2\right)+f\left(4\right)=\frac{ln 2}{2}+\frac{ln ⁡ 4}{4}=ln2$