The value of the integral $\int_0^1\frac{x^α-1}{ln\, x}dx$, is |
$ln\,α$ $2ln (α + 1)$ $3\,ln\,α$ none of these |
none of these |
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, $I(α) = ln |α + 1|$ |