Nice Integral

Put $x+1 = t$ to get the integral as :

$\displaystyle \int_{0}^{1} \frac{x^2-1}{\ln(x)} dx$

Let $I(a) = \int_{0}^{1} \frac{x^a-1}{\ln(x)} dx$

$\Rightarrow I'(a) = \int_{0}^{1} \frac{x^a \ln(x)}{\ln(x)} dx = \frac{1}{a+1}$

$\Rightarrow I(a) = \ln(a+1) + c$

Also, $I(0) = \displaystyle \int_{0}^{1} \frac{1-1}{\ln(x)} dx = 0$

Hence, $I(a) = \ln(a+1)$

Hence, our answer is $\ln(2+1) = \ln(3) = 1.09861$