# How do you find the Integral of ln(2x+1)?

It is $\int \left(\ln \left(2 x + 1\right)\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left(2 x + 1\right)$
$\int \ln \left(2 x + 1\right) \mathrm{dx} = \int \left(x\right) ' \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - \int x \cdot \left(\ln \left(2 x + 1\right)\right) ' \mathrm{dx} = x \ln \left(2 x + 1\right) - \int \left(x \cdot \frac{2}{2 x + 1}\right) = x \ln \left(2 x + 1\right) - \int \frac{2 x + 1}{2 x + 1} - \frac{1}{2 x + 1} \mathrm{dx} = x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left(2 x + 1\right)$