# Question ff170

Feb 21, 2017

$3 \ln \left({x}^{2} + 2 x + 2\right) - 12 {\tan}^{-} 1 \left(x + 1\right) + C$

#### Explanation:

$\int \frac{6 x - 6}{{x}^{2} + 2 x + 2} \mathrm{dx}$

$= \int \frac{6 x - 6}{\left({x}^{2} + 2 x + 1\right) + 1} \mathrm{dx}$

$= \int \frac{6 x - 6}{{\left(x + 1\right)}^{2} + 1} \mathrm{dx}$

In order to make the denominator resemble the trig identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$, let $\tan \theta = x + 1$.

This also imply that ${\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta = \mathrm{dx}$ and $x = \tan \theta - 1$. Substituting these in gives:

$= \int \frac{6 \left(\tan \theta - 1\right) - 6}{{\tan}^{2} \theta + 1} \left({\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta\right)$

$= \int \frac{6 \tan \theta - 12}{\sec} ^ 2 \theta \left({\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta\right)$

$= \int \left(6 \tan \theta - 12\right) d \theta$

Both of these are standard integrals. If you forget the integration of $\tan \theta$, recall that $\int \tan \theta \textcolor{w h i t e}{.} d \theta = \int \sin \frac{\theta}{\cos} \theta d \theta$, then use the substitution $u = \cos \theta$.

$= - 6 \ln \left\mid \cos \right\mid \theta - 12 \theta + C$

Our original was substitution was $\tan \theta = x + 1$, so $\theta = {\tan}^{-} 1 \left(x + 1\right)$.

Furthermore, if $\tan \theta = x + 1$, this is a triangle where the side opposite $\theta$ is $x + 1$ and the side adjacent is $1$, so the hypotenuse is $\sqrt{{\left(x + 1\right)}^{2} + 1} = \sqrt{{x}^{2} + 2 x + 2}$.

Then, $\cos \theta = \text{adjacent"/"hypotenuse} = \frac{1}{\sqrt{{x}^{2} + 2 x + 2}}$. So:

$= - 6 \ln \left\mid \frac{1}{\sqrt{{x}^{2} + 2 x + 2}} \right\mid - 12 {\tan}^{-} 1 \left(x + 1\right)$

Using $\frac{1}{\sqrt{{x}^{2} + 2 x + 2}} = {\left({x}^{2} + 2 x + 2\right)}^{- \frac{1}{2}}$ and the log rule $\ln \left({a}^{b}\right) = b \ln \left(a\right)$, this becomes:

= color(blue)(3ln(x^2+2x+2)-12tan^-1(x+1)+C#

The absolute value bars aren't necessary because ${x}^{2} + 2 x + 2 > 0$ for all real values of $x$.