The value of the integral ∫_0^1(x^α-1)/(ln x)dx, is

Let $I(α) =\int_0^1\frac{x^α-1}{ln\, x}dx$  .....(i)

Differentiating wrt α

$∴I’(α) =\int_0^1\frac{x^α.ln\, x}{ln\, x}dx=\int_0^1x^α\,dx$

$=\left[\frac{x^{α+1}}{α+1}\right]_0^1=\frac{1}{(α+1)}$

Now, $I(α) =\int\frac{dα}{(α+1)}= ln |α + 1| + c$

Put $α = 0$, then $I(0) = ln 1 + c = 0$ [from Eq. (i)]

$⇒ 0 + c = 0 ∴ c = 0$ Hence, $I(α) = ln |α + 1|$