# How do you find the integral ln(x^2+4)?

Oct 18, 2015

$I = x \ln \left({x}^{2} + 4\right) - 2 x + 4 \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

$I = \int \ln \left({x}^{2} + 4\right) \mathrm{dx}$

$u = \ln \left({x}^{2} + 4\right) \implies \mathrm{du} = \frac{1}{{x}^{2} + 4} \cdot 2 x \cdot \mathrm{dx} = \frac{2 x}{{x}^{2} + 4} \mathrm{dx}$

$\mathrm{dv} = \mathrm{dx} \implies v = x$

$I = x \ln \left({x}^{2} + 4\right) - \int x \frac{2 x}{{x}^{2} + 4} \mathrm{dx}$

$I = x \ln \left({x}^{2} + 4\right) - 2 \int {x}^{2} / \left({x}^{2} + 4\right) \mathrm{dx}$

$I = x \ln \left({x}^{2} + 4\right) - 2 \int \frac{{x}^{2} + 4 - 4}{{x}^{2} + 4} \mathrm{dx}$

$I = x \ln \left({x}^{2} + 4\right) - 2 \int \left(1 - \frac{4}{{x}^{2} + 4}\right) \mathrm{dx}$

$I = x \ln \left({x}^{2} + 4\right) - 2 \left(x - 4 \cdot \frac{1}{2} \arctan \left(\frac{x}{2}\right)\right) + C$

$I = x \ln \left({x}^{2} + 4\right) - 2 x + 4 \arctan \left(\frac{x}{2}\right) + C$