# What is the antiderivative of  ln(x^2 + 2x + 2)?

Apr 9, 2018

$x \ln \left({x}^{2} + 2 x + 2\right) - 2 x + \ln \left({x}^{2} + 2 x + 2\right) + 2 {\tan}^{-} 1 x + C$

#### Explanation:

The antiderivative of $\ln x$ can be easily found out by integration by parts to be

$\int \ln x \mathrm{dx} = \ln x - \int \left(\frac{d}{\mathrm{dx}} \left(\ln x\right) \times \int 1 \cdot \mathrm{dx}\right) \mathrm{dx}$
$q \quad = x \ln x - x + C$

The antiderivative of the given function is

$\int \ln \left({x}^{2} + 2 x + 2\right) \mathrm{dx} = \int \ln \left[\left({x}^{2} + 2 x + 2\right)\right] \cdot 1 \mathrm{dx}$

$q \quad = \ln \left({x}^{2} + 2 x + 2\right) \int 1 \cdot \mathrm{dx}$
$q \quad - \int \left[\frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 2 x + 2\right) \int 1. \mathrm{dx}\right] \mathrm{dx}$
$q \quad = x \ln \left({x}^{2} + 2 x + 2\right) - \int \frac{x \left(2 x + 2\right) \mathrm{dx}}{{x}^{2} + 2 x + 2}$

Now

$\frac{2 x \left(x + 1\right)}{{x}^{2} + 2 x + 2} = 2 \frac{{x}^{2} + 2 x + 2 - x - 2}{{x}^{2} + 2 x + 2}$
$q \quad = 2 - \frac{2 x + 4}{{x}^{2} + 2 x + 2}$
$q \quad = 2 - \frac{2 x + 2}{{x}^{2} + 2 x + 2} - \frac{2}{{x}^{2} + 2 x + 2}$

and so

$\int \frac{x \left(2 x + 2\right) \mathrm{dx}}{{x}^{2} + 2 x + 2}$
$q \quad = \int \left[2 - \frac{2 x + 2}{{x}^{2} + 2 x + 2} - \frac{2}{{\left(x + 1\right)}^{2} + 1}\right] \mathrm{dx}$
$q \quad = 2 x - \ln \left({x}^{2} + 2 x + 2\right) - 2 {\tan}^{-} 1 x + C$

Thus the required antiderivative is

$x \ln \left({x}^{2} + 2 x + 2\right) - 2 x + \ln \left({x}^{2} + 2 x + 2\right) + 2 {\tan}^{-} 1 x + C$