Past exam of the options I course of the University of Oxford / 2008 / D / D2 / a / Solution

The integrand is positive and continuous on the interval in question. Substitute $t=\log (1+x)$ to obtainIf $\alpha \ge 1$, the exponential factor is at most one and the comparison integral of $t^{-2}$ converges. If $\alpha <1$, exponential growth dominates $t^2$, so the integrand eventually exceeds one and the integral diverges. ThusThe logarithmic square is what makes the boundary value $\alpha =1$ integrable.