How do you evaluate the integral int x^2/(4+x^2)^2dx?

1 Answer
Aug 12, 2017

intcolor(white)(l)(x^2)/((4+x^2)^2)color(white)(l)dx = color(blue)(1/4(tan^-1[x/2] - (2x)/(x^2+4)) + C

Explanation:

We're asked to find the integral

$\int \textcolor{w h i t e}{l} \frac{{x}^{2}}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

Let's split this up into partial fractions:

$= \int \textcolor{w h i t e}{l} \frac{1}{{x}^{2} + 4} - \frac{4}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

$= \int \textcolor{w h i t e}{l} \frac{1}{{x}^{2} + 4} \textcolor{w h i t e}{l} \mathrm{dx} - 4 \int \textcolor{w h i t e}{l} \frac{1}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

$= \int \textcolor{w h i t e}{l} \frac{1}{4 \left(\frac{{x}^{2}}{4} + 1\right)} \textcolor{w h i t e}{l} \mathrm{dx} - 4 \int \textcolor{w h i t e}{l} \frac{1}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

$= \frac{1}{4} \int \textcolor{w h i t e}{l} \frac{1}{4 \left(\frac{{x}^{2}}{4} + 1\right)} \textcolor{w h i t e}{l} \mathrm{dx} - 4 \int \textcolor{w h i t e}{l} \frac{1}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

For the integrand $\frac{1}{\frac{{x}^{2}}{4} + 1}$, let's make $u = \frac{x}{2}$ and $\mathrm{dx} = 2 \textcolor{w h i t e}{l} \mathrm{du}$:

$= \frac{1}{2} \int \textcolor{w h i t e}{l} \frac{1}{{u}^{2} + 1} \textcolor{w h i t e}{l} \mathrm{du} - 4 \int \textcolor{w h i t e}{l} \frac{1}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

The integral of $\frac{1}{{u}^{2} + 1}$ is ${\tan}^{-} 1 u$:

$= \frac{1}{2} {\tan}^{-} 1 u - 4 \int \textcolor{w h i t e}{l} \frac{1}{{\left({x}^{2} + 4\right)}^{2}} \textcolor{w h i t e}{l} \mathrm{dx}$

For the integrand $\frac{1}{{\left({x}^{2} + 4\right)}^{2}}$, let's make $x = 2 \tan s$ and $\mathrm{dx} = 2 {\sec}^{2} s \textcolor{w h i t e}{l} \mathrm{ds}$.

Then ${\left({x}^{2} + 4\right)}^{2} = {\left(4 {\tan}^{2} s + 4\right)}^{2} = 16 {\sec}^{4} s$ and $s = {\tan}^{-} 1 \left[\frac{x}{2}\right]$:

$= \frac{1}{2} {\tan}^{-} 1 u - 8 \int \textcolor{w h i t e}{l} \frac{1}{16} {\cos}^{2} s \textcolor{w h i t e}{l} \mathrm{ds}$

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{2} \int \textcolor{w h i t e}{l} {\cos}^{2} s \textcolor{w h i t e}{l} \mathrm{ds}$

We can write ${\cos}^{2} s$ as $\frac{1}{2} \cos \left[2 s\right] + \frac{1}{2}$:

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{2} \int \textcolor{w h i t e}{l} \frac{1}{2} \cos \left[2 s\right] + \frac{1}{2} \textcolor{w h i t e}{l} \mathrm{ds}$

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{4} \int \textcolor{w h i t e}{l} \cos \left[2 s\right] \textcolor{w h i t e}{l} \mathrm{ds} - \frac{1}{4} \int \textcolor{w h i t e}{l} 1 \textcolor{w h i t e}{l} \mathrm{ds}$

For the integrand $\cos \left[2 s\right]$, let's make $p = 2 s$ and $\mathrm{ds} = \frac{1}{2} \mathrm{dp}$:

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{8} \int \textcolor{w h i t e}{l} \cos p \textcolor{w h i t e}{l} \mathrm{dp} - \frac{1}{4} \int \textcolor{w h i t e}{l} 1 \textcolor{w h i t e}{l} \mathrm{ds}$

The integral of $\cos p$ is $\sin p$:

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{8} \sin p - \frac{1}{4} \int \textcolor{w h i t e}{l} 1 \textcolor{w h i t e}{l} \mathrm{ds}$

The integral of $1$ is $s$:

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{8} \sin p - \frac{1}{4} s + C$

Substitute back in $p = 2 s$:

$= \frac{1}{2} {\tan}^{-} 1 u - \frac{1}{8} \sin \left[2 s\right] - \frac{1}{4} s + C$

Substitute back in $s = {\tan}^{-} 1 \left[\frac{x}{2}\right]$:

$= \frac{\left({x}^{2} + 4\right) \left(2 {\tan}^{-} 1 u - {\tan}^{-} 1 \left[\frac{x}{2}\right]\right) - 2 x}{4 \left({x}^{2} + 4\right)} + C$

Substitute back in $u = \frac{x}{2}$:

$= \frac{\left({x}^{2} + 4\right) {\tan}^{-} 1 \left[\frac{x}{2}\right] - 2 x}{4 \left({x}^{2} + 4\right)} + C$

Or

color(blue)(ulbar(|stackrel(" ")(" "I = 1/4(tan^-1[x/2] - (2x)/(x^2+4)) + C" ")|)