# How to intergrate this? The answer should be -2 ln(x^2 +1). I thought it was -4x*ln(x^2 +1), can someone explain to me how to get to the answer? It’s not about filling in the 1 and 0, but I didn’t know how to get rid of that :).

## ${\int}_{0}^{1} \left(\frac{- 4 x}{{x}^{2} + 1}\right) \mathrm{dx}$

Jun 17, 2018

the result is given by $- 2 \ln \left(2\right)$
$- 2 {\int}_{0}^{1} \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$
$- {\left[- 2 \ln \left({x}^{2} + 1\right)\right]}_{0}^{1} = - 2 \ln \left(2\right)$