Question

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

Answer 1

The answer is`intx^4/(x^4-1)=x-1/2arctan(x)-1/4ln(x+1)+1/4ln(x-1)`

Explanation:

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

`(x^4-1)sqrt(x^4)`

=`1+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=(x^2+1)(x^2-1)`

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

`x^2-1=(x+1)(x-1)`

So, overall the denominator is:

`x^4-1=(x^2+1)(x+1)(x-1)`

Then, we do partial fraction:

`1/((x^2+1)(x+1)(x-1))=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)`

`1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)`

subtitute `x=1` into the equation

`1=4D`
`D=1/4`

subtitute `x=-1` into the equation

`1=-4C`
`C=-1/4`

compare the coefficient of `x^3`on both side

`0=A+C+D`
`A=-C-D`
`=1/4-1/4`
`A=0`

compare the constant number on both side

`1=-B-C+D`
`B=-1-C+D`
`=-1+1/4+1/4`
`B=-1/2`

so

`intx^4/(x^4-1)=int(1-1/(2(x^2+1))-1/(4(x+1))+1/(4(x-1)))`

`intx^4/(x^4-1)=x-int1/2*1/(x^2+1)-int1/4*1/(x+1)+int1/4*1/(x-1)`

`intx^4/(x^4-1)=x-1/2int1/(x^2+1)-1/4int1/(x+1)+1/4int1/(x-1)`

use formula :

`int(f'(x))/(f(x))=Inf(x)`

`intx^4/(x^4-1)=x-1/2arctan(x)-1/4ln(x+1)+1/4ln(x-1)`

Answer 2

`I=x+1/4ln|(x-1)/(x+1)|-1/2tan^-1x+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)#

`=>x^4/(x^4-1)=1+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/(x^4-1)=1+1/2[color(blue)(2)/((x^2-1)(x^2+1))]`

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

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

Integrating each term we get

`I=intx^4/(x^4-1)dx=int1dx+1/2int[color(red)(1/(x^2-1))-color(brown)(1/(x^2+1))]dx`

`I=x+1/2[color(red)(1/2ln|(x-1)/(x+1)|)-color(brown)(tan^-1x)]+c`

`I=x+1/4ln|(x-1)/(x+1)|-1/2tan^-1x+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`