# Stuck on this integral (using partial fraction decomposition)

• QaH
In summary: I don't know if there is a comprehensive list anywhere.In summary, the integrand \int\frac{x^2}{\sqrt{x^2+4}}dx can be solved by using the substitution x=2tan\theta and dx=2sec^2\theta d\theta. After performing partial fraction decomposition and using a difference of perfect squares, the integral can be simplified to \frac{1}{(1-u^2)^2}. This can then be further simplified by using the substitution u=1 and applying the Pythagorean identity. Alternatively, using the substitution x = 2sinh(t) simplifies the integral further.
QaH

## Homework Statement

$$\int\frac{x^2}{\sqrt{x^2+4}}dx$$

n/a

## The Attempt at a Solution

Letting $$x=2tan\theta$$ and $$dx=2sec^2\theta d\theta$$
$$\int\frac{x^2}{\sqrt{x^2+4}}dx=\int\frac{4tan^2\theta}{\sqrt{4+4tan^2\theta}}2sec^2\theta d\theta=\int\frac{8tan^2\theta sec^2\theta}{\sqrt{4(1+tan^2\theta)}}d\theta=4\int\frac{tan^2\theta sec^2\theta}{sec\theta}d\theta=$$$$4\int tan^2\theta sec\theta d\theta=4\int(sec^2\theta -1)sec\theta d\theta=4\int (sec^3\theta-sec\theta)d\theta=$$$$4\int sec^3\theta d\theta-4\int sec\theta d\theta=-4\int\frac{cos\theta}{cos^2\theta}d\theta+4\int\frac{cos\theta}{cos^4\theta}d\theta=$$$$-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(cos^2\theta)^2}d\theta=-4\int\frac{cos\theta}{1-sin^2\theta}d\theta+4\int\frac{cos\theta}{(1-sin^2\theta)^2}d\theta$$
now letting $$u=sin\theta$$and$$du=cos\theta d\theta$$$$-4\int\frac{1}{1-u^2}du+4\int\frac{1}{(1-u^2)^2}du=-4\int\frac{1}{(1-u)(1+u)}du+4\int\frac{1}{(1-u^2)^2}du$$
using partial fraction decomposition I get
$$\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$$ multipying both sides by $$1-u^2$$$$1=A(1+u)+B(1-u)$$ Letting u=1 we get that $$A=\frac{1}{2}$$ now letting u=-1 we get that $$B=\frac{1}{2}$$ Then $$\frac{1}{(1-u)(1+u)}=\frac{1}{2(1-u)}+\frac{1}{2(1+u)}$$
and $$-4\int\frac{1}{(1-u)(1+u)}du$$ becomes $$-4\int\frac{1}{2}(\frac{1}{1-u}+\frac{1}{1+u})du=-2(\ln\mid1+u\mid-\ln\mid1-u\mid+c$$note that $$u=sin\theta=\frac{x}{\sqrt{x^2+4}}$$$$-2(\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid-\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid)+c=-2\ln\mid1+\frac{x}{\sqrt{x^2+4}}\mid+2\ln\mid1-\frac{x}{\sqrt{x^2+4}}\mid+c=\ln\mid(1-\frac{x}{\sqrt{x^2+4}})^2\mid-\ln\mid(1+\frac{x}{\sqrt{x^2+4}})^{2}\mid+c=\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+c$$
and our original integral becomes $$\ln\mid\frac{(1-\frac{x}{\sqrt{x^2+4}})^2}{(1+\frac{x}{\sqrt{x^2+4}})^2}\mid+4\int\frac{1}{(1-u^2)^2}du$$
now this is where I am stuck because I can't seem to figue out partial fraction decomposition on $$\frac{1}{(1-u^2)^2}=\frac{A}{(1-u^2)}+\frac{B}{(1-u^2)^2}$$ multiplying through I get that $$1=A(1-u^2)+B$$
Letting u=1 B=1 then$$1=A(1-u^2)+1$$ then for any u A=0 and I am left with $$\frac{1}{(1-u^2)^2}=0+\frac{1}{(1-u^2)^2}$$ and I still can't integrate.
Please don't just say I am not using the correct partial fraction method because I have searched and searched but can't find a method on this type of fraction.

Have you tried the complete decomposition?
$$\dfrac{1}{(1-u^2)^2}=\dfrac{A}{1-u}+\dfrac{B}{1-u}+\frac{C}{1+u}+\frac{D}{1+u}$$
or
$$\dfrac{1}{(1-u^2)^2}=\dfrac{Au+B}{(1+u)^2}+\dfrac{Cu+D}{(1-u)^2}$$

A key idea when doing partial fraction decomposition, is that to solve problems of this type, you have to make sure the denominator is fully factored. Also, you have to make sure that if the denominators contain quadratics, i.e. X^2+1, you need to do something to preform a little "fix." Now try to think as to why need to preform this "fix."

What you have in those parenthesis is a difference of perfect squares...

The mathematical reasoning as to why and how partial fraction decomposition work is really neat. At this point you should have enough background to attempt it! give it a try.

Theres also a really cool shortcut when working with denominators that are "linear." Youtube it!

QaH said:

## Homework Statement

$$\int\frac{x^2}{\sqrt{x^2+4}}dx$$

n/a

## The Attempt at a Solution

Letting $$x=2tan\theta$$ and $$dx=2sec^2\theta d\theta$$

$$\vdots\\ \vdots$$

Letting u=1 B=1 then$$1=A(1-u^2)+1$$ then for any u A=0 and I am left with $$\frac{1}{(1-u^2)^2}=0+\frac{1}{(1-u^2)^2}$$ and I still can't integrate.
Please don't just say I am not using the correct partial fraction method because I have searched and searched but can't find a method on this type of fraction.

(1)
$$\frac{1}{(1-u^2)^2} = \left[ \frac{1}{2}\frac{1}{1+u} + \frac{1}{2} \frac{1}{1-u} \right]^2.$$
Expand ##[ \cdots ]^2## and re-apply partial fractions to some of the terms; others are already in simplest form.
(2) You could use ##x = 2 \sinh(t)## instead of your ## x = 2 \tan \theta##. I think it makes things a lot simpler.

BTW: do not write ##tan \theta##; it is hard to read and looks ugly: ##\tan \theta## looks a lot better. LaTeX is designed to produce nice output if you put a "\" in front of most standard functions, so instead of writing "sin" or "tan" you should use "\ sin" or "\ tan"---but leave out the space between '\' and sin or tan. This applies as well to most other elementary functions cos, arcsin, arccos, arctan, sinh, cosh, tanh, exp, ln, log, max, min, lim, sup, inf, gcd, and some others.

Last edited:
QaH and epenguin

## 1. What is partial fraction decomposition?

Partial fraction decomposition is a mathematical technique used to break down a complicated rational function into simpler fractions. This allows for easier integration and manipulation of the function.

## 2. When should I use partial fraction decomposition?

Partial fraction decomposition is typically used when integrating rational functions in calculus. It can also be used in other areas of mathematics and engineering to simplify complex equations.

## 3. How do I perform partial fraction decomposition?

To perform partial fraction decomposition, you first need to factor the denominator of the rational function into linear and quadratic factors. Then, you set up a system of equations and solve for the unknown coefficients. Finally, you integrate each term separately.

## 4. What are the benefits of using partial fraction decomposition?

Partial fraction decomposition allows for easier integration and manipulation of rational functions. It can also help identify important features of the function, such as asymptotes and roots.

## 5. Are there any limitations to using partial fraction decomposition?

Partial fraction decomposition can only be used for rational functions with distinct linear and quadratic factors in the denominator. It also does not work for functions with repeated factors or complex roots. In these cases, alternative techniques may need to be used.

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