How do you evaluate the integral $\int \frac{x^{2}}{{(4 + x^{2})}^{2}} d x$ $int x^2/(4+x^2)^2dx$ ?

Question

How do you evaluate the integral $\int \frac{x^{2}}{{(4 + x^{2})}^{2}} d x$ $int x^2/(4+x^2)^2dx$ ?

1 Answer

Impact of this question

18091 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-12T01:37:21

$\int l \frac{x^{2}}{{(4 + x^{2})}^{2}} l d x = \frac{1}{4} ({tan}^{- 1} [\frac{x}{2}] - \frac{2 x}{x^{2} + 4}) + C$ $intcolor(white)(l)(x^2)/((4+x^2)^2)color(white)(l)dx = color(blue)(1/4(tan^-1[x/2] - (2x)/(x^2+4)) + C$

Explanation:

We're asked to find the integral

$\int l \frac{x^{2}}{{(x^{2} + 4)}^{2}} l d x$ $intcolor(white)(l)(x^2)/((x^2+4)^2)color(white)(l)dx$

Let's split this up into partial fractions:

$= \int l \frac{1}{x^{2} + 4} - \frac{4}{{(x^{2} + 4)}^{2}} l d x$ $= intcolor(white)(l)1/(x^2+4) - 4/((x^2+4)^2)color(white)(l)dx$

$= \int l \frac{1}{x^{2} + 4} l d x - 4 \int l \frac{1}{{(x^{2} + 4)}^{2}} l d x$ $= intcolor(white)(l)1/(x^2+4)color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx$

$= \int l \frac{1}{4 (\frac{x^{2}}{4} + 1)} l d x - 4 \int l \frac{1}{{(x^{2} + 4)}^{2}} l d x$ $= intcolor(white)(l)1/(4((x^2)/4+1))color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx$

$= \frac{1}{4} \int l \frac{1}{4 (\frac{x^{2}}{4} + 1)} l d x - 4 \int l \frac{1}{{(x^{2} + 4)}^{2}} l d x$ $= 1/4intcolor(white)(l)1/(4((x^2)/4+1))color(white)(l)dx - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx$

For the integrand $\frac{1}{\frac{x^{2}}{4} + 1}$ $1/((x^2)/4+1)$ , let's make $u = \frac{x}{2}$ $u = x/2$ and $d x = 2 l d u$ $dx = 2color(white)(l)du$ :

$= \frac{1}{2} \int l \frac{1}{u^{2} + 1} l d u - 4 \int l \frac{1}{{(x^{2} + 4)}^{2}} l d x$ $= 1/2intcolor(white)(l)1/(u^2+1)color(white)(l)du - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx$

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

$= \frac{1}{2} {tan}^{- 1} u - 4 \int l \frac{1}{{(x^{2} + 4)}^{2}} l d x$ $= 1/2tan^-1u - 4intcolor(white)(l)1/((x^2+4)^2)color(white)(l)dx$

For the integrand $\frac{1}{{(x^{2} + 4)}^{2}}$ $1/((x^2+4)^2)$ , let's make $x = 2 tan s$ $x = 2tans$ and $d x = 2 {sec}^{2} s l d s$ $dx = 2sec^2scolor(white)(l)ds$ .

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

$= \frac{1}{2} {tan}^{- 1} u - 8 \int l \frac{1}{16} {cos}^{2} s l d s$ $= 1/2tan^-1u - 8intcolor(white)(l)1/16cos^2scolor(white)(l)ds$

$= \frac{1}{2} {tan}^{- 1} u - \frac{1}{2} \int l {cos}^{2} s l d s$ $= 1/2tan^-1u - 1/2intcolor(white)(l)cos^2scolor(white)(l)ds$

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

$= \frac{1}{2} {tan}^{- 1} u - \frac{1}{2} \int l \frac{1}{2} cos [2 s] + \frac{1}{2} l d s$ $= 1/2tan^-1u - 1/2intcolor(white)(l)1/2cos[2s] + 1/2color(white)(l)ds$

$= \frac{1}{2} {tan}^{- 1} u - \frac{1}{4} \int l cos [2 s] l d s - \frac{1}{4} \int l 1 l d s$ $= 1/2tan^-1u - 1/4intcolor(white)(l)cos[2s]color(white)(l)ds - 1/4intcolor(white)(l)1color(white)(l)ds$

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

$= \frac{1}{2} {tan}^{- 1} u - \frac{1}{8} \int l cos p l d p - \frac{1}{4} \int l 1 l d s$ $= 1/2tan^-1u - 1/8intcolor(white)(l)cospcolor(white)(l)dp - 1/4intcolor(white)(l)1color(white)(l)ds$

The integral of $cos p$ $cosp$ is $sin p$ $sinp$ :

$= \frac{1}{2} {tan}^{- 1} u - \frac{1}{8} sin p - \frac{1}{4} \int l 1 l d s$ $= 1/2tan^-1u - 1/8sinp - 1/4intcolor(white)(l)1color(white)(l)ds$

The integral of $1$ $1$ is $s$ $s$ :

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

Substitute back in $p = 2 s$ $p = 2s$ :

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

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

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

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

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

Or

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

Science

Math

Humanities

... and beyond

How do you evaluate the integral $\int \frac{x^{2}}{{(4 + x^{2})}^{2}} d x$ $int x^2/(4+x^2)^2dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate the integral int x^2/(4+x^2)^2dx∫x2(4+x2)2dxint x^2/(4+x^2)^2dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate the integral $\int \frac{x^{2}}{{(4 + x^{2})}^{2}} d x$ $int x^2/(4+x^2)^2dx$ ?