# How do you integrate int x^4/(x^4-1) using partial fractions?

Sep 26, 2016

The answer is$\int {x}^{4} / \left({x}^{4} - 1\right) = x - \frac{1}{2} \arctan \left(x\right) - \frac{1}{4} \ln \left(x + 1\right) + \frac{1}{4} \ln \left(x - 1\right)$

#### Explanation:

Since the denominator is the same as the numerator, you need to do long division first:

$\left({x}^{4} - 1\right) \sqrt{{x}^{4}}$

=$1 + \frac{1}{{x}^{4} - 1}$

Then, we started to do partial fraction.
Since the denominator is not a linear function, we need to check it if it is reducible. Therefore, the denominator is reduced to:

${x}^{4} - 1 = \left({x}^{2} + 1\right) \left({x}^{2} - 1\right)$

These factors are also not linear function. Therefore, we check again whether these funtion can be reduce and we find out:

${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$

So, overall the denominator is:

${x}^{4} - 1 = \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)$

Then, we do partial fraction:

$\frac{1}{\left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)} = \frac{A x + B}{{x}^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$

$1 = \left(A x + B\right) \left(x + 1\right) \left(x - 1\right) + C \left({x}^{2} + 1\right) \left(x - 1\right) + D \left({x}^{2} + 1\right) \left(x + 1\right)$

subtitute $x = 1$ into the equation

$1 = 4 D$
$D = \frac{1}{4}$

subtitute $x = - 1$ into the equation

$1 = - 4 C$
$C = - \frac{1}{4}$

compare the coefficient of ${x}^{3}$on both side

$0 = A + C + D$
$A = - C - D$
$= \frac{1}{4} - \frac{1}{4}$
$A = 0$

compare the constant number on both side

$1 = - B - C + D$
$B = - 1 - C + D$
$= - 1 + \frac{1}{4} + \frac{1}{4}$
$B = - \frac{1}{2}$

so

$\int {x}^{4} / \left({x}^{4} - 1\right) = \int \left(1 - \frac{1}{2 \left({x}^{2} + 1\right)} - \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x - 1\right)}\right)$

$\int {x}^{4} / \left({x}^{4} - 1\right) = x - \int \frac{1}{2} \cdot \frac{1}{{x}^{2} + 1} - \int \frac{1}{4} \cdot \frac{1}{x + 1} + \int \frac{1}{4} \cdot \frac{1}{x - 1}$

$\int {x}^{4} / \left({x}^{4} - 1\right) = x - \frac{1}{2} \int \frac{1}{{x}^{2} + 1} - \frac{1}{4} \int \frac{1}{x + 1} + \frac{1}{4} \int \frac{1}{x - 1}$

use formula :

$\int \frac{f ' \left(x\right)}{f \left(x\right)} = I n f \left(x\right)$

$\int {x}^{4} / \left({x}^{4} - 1\right) = x - \frac{1}{2} \arctan \left(x\right) - \frac{1}{4} \ln \left(x + 1\right) + \frac{1}{4} \ln \left(x - 1\right)$

Jun 3, 2018

$I = x + \frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \frac{1}{2} {\tan}^{-} 1 x + c$

#### Explanation:

Sometimes we can obtain Partial fraction with simple adjustment methods, without using A, B, C..etc.The advantage of this method is : no need to solve for A,B ,C...etc

Here,

x^4/(x^4-1)=((x^4color(violet)(-1))+color(violet)(1))/(x^4-1)= (x^4-1)/(x^4-1)+1/(x^4-1)

$\implies {x}^{4} / \left({x}^{4} - 1\right) = 1 + \frac{1}{{x}^{4} - 1}$

=>x^4/(x^4-1)=1+1/((x^2-1)(x^2+1)

Now,

color(blue)((x^2+1)-(x^2-1)=2

So,

${x}^{4} / \left({x}^{4} - 1\right) = 1 + \frac{1}{2} \left[\frac{\textcolor{b l u e}{2}}{\left({x}^{2} - 1\right) \left({x}^{2} + 1\right)}\right]$

${x}^{4} / \left({x}^{4} - 1\right) = 1 + \frac{1}{2} \left[\frac{\textcolor{b l u e}{\left(\left({x}^{2} + 1\right) - \left({x}^{2} - 1\right)\right)}}{\left({x}^{2} - 1\right) \left({x}^{2} + 1\right)}\right]$

x^4/(x^4-1)=1+1/2[(x^2+1)/((x^2-1)(x^2+1))-(x^2-1)/((x^2-1) (x^2+1))]

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

Integrating each term we get

$I = \int {x}^{4} / \left({x}^{4} - 1\right) \mathrm{dx} = \int 1 \mathrm{dx} + \frac{1}{2} \int \left[\textcolor{red}{\frac{1}{{x}^{2} - 1}} - \textcolor{b r o w n}{\frac{1}{{x}^{2} + 1}}\right] \mathrm{dx}$

$I = x + \frac{1}{2} \left[\textcolor{red}{\frac{1}{2} \ln | \frac{x - 1}{x + 1} |} - \textcolor{b r o w n}{{\tan}^{-} 1 x}\right] + c$

$I = x + \frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \frac{1}{2} {\tan}^{-} 1 x + c$

Note:

color(red)((1)int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c

color(brown)((2)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c