Question

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

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

Answer 1

The correct option is A $\log (\alpha +1)$

$I\left(\alpha \right)={\int }_0^1\frac{x^{\alpha }-1}{\text{log}x}dx$

Differentiating with respect to $\alpha ,$ we get$,$

$I^{\prime }\left(\alpha \right)={\int }_0^1{x}^{\alpha }dx=\frac{1}{\alpha +1}$

We need to solve this simple differential equation $:I^{\prime }\left(\alpha \right)=\frac{1}{\alpha +1}$

On solving, we get,

$I\left(\alpha \right)=\text{log}(\alpha +1)+C$

Now, from the initial equation, we can find that $I\left(0\right)=0$

$I\left(0\right)=\text{log}\left(1\right)+C=0$ $\Rightarrow C=0$

So, $I\left(\alpha \right)=\text{log}(\alpha +1)$