Question

The value of the integral $\displaystyle\int_{0}^{1}\frac{x^b-1}{\log\,x}dx$ is

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

Answer 1

Answer: (D)

Let $I \left(b\right)=\int\limits_{0}^{1} \frac{x^{b}-1}{log\,x}dx$
$\therefore I' \left(b\right)=\int\limits_{0}^{1} \frac{x^{b}-log\,x}{log\,x}dx=\int\limits_{0}^{1}x^{b} dx$
$=\left|\frac{x^{b+1}}{b+1}\right|_{0}^{1}=\frac{1}{b+1}$
$\therefore I\left(b\right)=log\left(b+1\right)+C$
When $b - 0 .1\left(b\right) = 0 \therefore c = 0$
Hence $I\left(b\right) = log \left(b + 1\right)$