Question

Evaluate: ${\int }_1^e\frac{\ln x}{x^2}dx$

• A. 1 $-\frac{2}{e}$
• B. $\frac{2}{e}-1$
• C. $\frac{e-1}{e}$
• D. $1+\frac{2}{e}$

Answer 1

Answer: (A)

1 $-\frac{2}{e}$

${\int }_1^e\frac{\ln x}{x^2}dx$

Let $z=\ln x$ $⟹dz=\frac{1}{x}$ and $x=e^z$

When $x=1,z=0$ and when $x=e,z=1$, hence integration becomes:-

${\int }_1^e\frac{1}{x}\ln x\frac{dx}{x}$

$={\int }_0^1{e}^{-z}.zdz$

Using integration by parts:-

$={[-z{e}^{-z}-e^{-z}]}_0^1$

$=(-e^{-1}-e^{-1})-(-1)=-2e^{-1}+1$

$=1-\frac{2}{e}$

Hence, answer is option-(A).