Question

Which of the following integrals is unbounded?

• A. ${\int }_0^{\pi /4}tanxdx$
• B. ${\int }_0^{\infty }\frac{1}{x^2+1}dx$
• C. ${\int }_0^{\infty }x{e}^{-x}dx$
• D. ${\int }_0^1\frac{1}{1-x}dx$

Answer 1

The correct option is D ${\int }_0^1\frac{1}{1-x}dx$

On integrating only option (d) does not have finite ans.
For $I={\int }_0^1\frac{dx}{1-x}={[-log(1-x\left)\right]}_0^1$
$={\left[log\left(\frac{1}{1-x}\right)\right]}_0^1=log\infty -log1=\infty$