Question #9bf04

Jun 17, 2017

$\frac{- 3 x - 1}{4 \left({x}^{2} + 2 x - 1\right)} + \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{x - \sqrt{2} + 1}{x + \sqrt{2} + 1} \right\mid + C$

Explanation:

Rewriting the denominator:

$\int \frac{3 {x}^{2} + 5 x - 1}{{x}^{2} + 2 x - 1} ^ 2 \mathrm{dx} = \int \frac{3 {x}^{2} + 5 x - 1}{\left({x}^{2} + 2 x + 1\right) - 2} ^ 2 \mathrm{dx} = \int \frac{3 {x}^{2} + 5 x - 1}{{\left(x + 1\right)}^{2} - 2} ^ 2 \mathrm{dx}$

Let $u = x + 1$, implying that $x = u - 1$ and $\mathrm{du} = \mathrm{dx}$.

$= \int \frac{3 {\left(u - 1\right)}^{2} + 5 \left(u - 1\right) - 1}{{u}^{2} - 2} ^ 2 \mathrm{du} = \int \frac{3 {u}^{2} - u - 3}{{u}^{2} - 2} ^ 2 \mathrm{du}$

Now let $u = \sqrt{2} \sec \theta$. This implies that ${u}^{2} - 2 = 2 {\sec}^{2} \theta - 2 = 2 {\tan}^{2} \theta$ and that $\mathrm{du} = \sqrt{2} \sec \theta \tan \theta d \theta$.

$= \int \frac{3 \left(2 {\sec}^{2} \theta\right) - \sqrt{2} \sec \theta - 3}{2 {\tan}^{2} \theta} ^ 2 \left(\sqrt{2} \sec \theta \tan \theta d \theta\right)$

$= \frac{\sqrt{2}}{4} \int \frac{6 {\sec}^{3} \theta - \sqrt{2} {\sec}^{2} \theta - 3 \sec \theta}{\tan} ^ 3 \theta d \theta$

Multiplying through by ${\cos}^{3} \frac{\theta}{\cos} ^ 3 \theta$:

$= \frac{1}{2 \sqrt{2}} \int \frac{6 - \sqrt{2} \cos \theta - 3 {\cos}^{2} \theta}{\sin} ^ 3 \theta d \theta$

$= \frac{3}{\sqrt{2}} \int {\csc}^{3} \theta d \theta - \frac{1}{2} \int \cot \theta {\csc}^{2} \theta d \theta - \frac{3}{2 \sqrt{2}} \int \frac{1 - {\sin}^{2} \theta}{\sin} ^ 3 \theta d \theta$

Note that $\frac{3}{\sqrt{2}} - \frac{3}{2 \sqrt{2}} = \frac{3}{2 \sqrt{2}}$:

$= \frac{3}{2 \sqrt{2}} \int {\csc}^{3} \theta d \theta - \frac{1}{2} \int \cot \theta {\csc}^{2} \theta d \theta + \frac{3}{2 \sqrt{2}} \int \csc \theta d \theta$

Find the method for integrating ${\csc}^{3} \theta$ here.

For the second integral, let $v = \cot \theta$ so $\mathrm{dv} = - {\csc}^{2} \theta d \theta$.

The integral of $\csc \theta$ is well known. For its derivation, see here.

$= \frac{3}{2 \sqrt{2}} \left(- \frac{1}{2}\right) \left(\cot \theta \csc \theta + \ln \left\mid \cot \theta + \csc \theta \right\mid\right) + \frac{1}{2} \int v \mathrm{dv} + \frac{3}{2 \sqrt{2}} \ln \left\mid \csc \theta - \cot \theta \right\mid$

Note that $\frac{1}{2} \int v \mathrm{dv} = \frac{1}{2} \left({v}^{2} / 2\right) = {v}^{2} / 4 = {\cot}^{2} \frac{\theta}{4}$.

$= - \frac{3}{4 \sqrt{2}} \cot \theta \csc \theta - \frac{3}{4 \sqrt{2}} \ln \left\mid \cot \theta + \csc \theta \right\mid + \frac{3}{4 \sqrt{2}} \ln \left\mid \csc \theta - \cot \theta \right\mid + \frac{1}{4} {\cot}^{2} \theta$

Our substitution was $u = \sqrt{2} \sec \theta$, implying that $\cos \theta = \frac{\sqrt{2}}{u}$, which is a right triangle where the side adjacent to $\theta$ is $\sqrt{2}$, the hypotenuse is $u$, and the side opposite $\theta$ is $\sqrt{{u}^{2} - 2}$.

Thus, $\csc \theta = \frac{u}{\sqrt{{u}^{2} - 2}}$ and $\cot \theta = \frac{\sqrt{2}}{\sqrt{{u}^{2} - 2}}$.

Also note that $- \frac{3}{4 \sqrt{2}} \ln \left\mid \cot \theta + \csc \theta \right\mid + \frac{3}{4 \sqrt{2}} \ln \left\mid \csc \theta - \cot \theta \right\mid = \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{\csc \theta - \cot \theta}{\cot \theta + \csc \theta} \right\mid = \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{1 - \cos \theta}{1 + \cos \theta} \right\mid$.

$= - \frac{3}{4 \sqrt{2}} \frac{\sqrt{2} u}{{u}^{2} - 2} + \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{1 - \frac{\sqrt{2}}{u}}{1 + \frac{\sqrt{2}}{u}} \right\mid + \frac{1}{4} \left(\frac{2}{{u}^{2} - 2}\right)$

$= - \frac{3}{4} \left(\frac{u}{{u}^{2} - 2}\right) + \frac{1}{2} \left(\frac{1}{{u}^{2} - 2}\right) + \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{u - \sqrt{2}}{u + \sqrt{2}} \right\mid$

$= \frac{- 3 u + 2}{4 \left({u}^{2} - 2\right)} + \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{u - \sqrt{2}}{u + \sqrt{2}} \right\mid$

With $u = x + 1$:

$= \frac{- 3 x - 1}{4 \left({x}^{2} + 2 x - 1\right)} + \frac{3}{4 \sqrt{2}} \ln \left\mid \frac{x - \sqrt{2} + 1}{x + \sqrt{2} + 1} \right\mid + C$