Question

Evaluate the integral

$\stackrel{1}{\underset{0}{\int }}\log \left(\begin{array}{c}\frac{1}{x}-1\end{array}\right)dx$

• A. $1$
• B. $0$
• C. $2$
• D. None of these

Answer 1

The correct option is B $0$

Let $I=\stackrel{1}{\underset{0}{\int }}\log \left(\begin{array}{c}\frac{1}{x}-1\end{array}\right)dx={\int }_0^1\log \left(\frac{1-x}{x}\right)dx$

$\Rightarrow I={\int }_0^1\left[\log \right(1-x)-\log x]dx$.......$\left(1\right)$

Now suing ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^1\left[\log \right(1-(1-x))-\log (1-x\left)\right]dx={\int }_0^1[\log x-\log (1-x\left)\right]dx$.....$\left(2\right)$

Now add $\left(1\right)$ and $\left(2\right)$, we get

$2I=0\Rightarrow I=0$