Question

The value of the integral ${\int }_0^1\frac{x^b-1}{\log x}dx$ is

• A. $\log b$
• B. $2\log (b+1)$
• C. $3\log b$
• D. none of these

Answer 1

Answer: (D)

Let $I\left(b\right)=\underset{0}{\overset{1}{\int }}\frac{x^b-1}{logx}dx$
$\therefore I^{\prime }\left(b\right)=\underset{0}{\overset{1}{\int }}\frac{x^b-logx}{logx}dx=\underset{0}{\overset{1}{\int }}{x}^{b}dx$
$={\left|\frac{x^{b+1}}{b+1}\right|}_0^1=\frac{1}{b+1}$
$\therefore I\left(b\right)=log\left(b+1\right)+C$
When $b-0.1\left(b\right)=0\therefore c=0$
Hence $I\left(b\right)=log\left(b+1\right)$