Question

How do you integrate `(6x)/(x^3-8)` using partial fractions?

Answer 1

The answer is `=ln(∣x-2∣)-1/2ln(∣x^2+2x+4∣)+4/sqrt3arctan((x+1)/sqrt3)+C`

Explanation:

The denominator is

`x^3-8=(x-2)(x^2+2x+4)`

Therefore,

`(6x)/(x^3-8)=(6x)/((x-2)(x^2+2x+4))`

`=A/(x-2)+(Bx+C)/(x^2+2x+4)`

`=(A(x^2+2x+4)+(Bx+C)(x-2))/((x-2)(x^2+2x+4))`

So,

`6x=A(x^2+2x+4)+(Bx+C)(x-2)`

Let `x=2`, `=>`, `12=12A`, `=>`, `A=1`

Let `x=0`, `=>`, `0=4A-2C`, `=>`, `C=2`

Coefficients of `x^2`, `=>`, `0=A+B`, `=>`, `B=-1`

Therefore,

`(6x)/(x^3-8)=1/(x-2)+(-x+2)/(x^2+2x+4)`

So,

`int(6xdx)/(x^3-8)=intdx/(x-2)-int((x-2)dx)/(x^2+2x+4)`

We integrate separately

`intdx/(x-2)=ln(∣x-2∣)`

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

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

Let `u=x^2+2x+4`, `=>`, `du=(2x+2)dx`

Therefore,

`int((x+2)dx)/(x^2+2x+4)=1/2int(du)/u=1/2lnu`

`=1/2ln(∣x^2+2x+4∣)`

`x^2+2x+4=x^2+2x+1+3=(x+1)^2+3`

`int(4dx)/(x^2+2x+4)=4int(dx)/((x+1)^2+3)`

`=4int(dx)/(3(((x+1)/sqrt3)^2+1))`

`=4/3int(dx)/((((x+1)/sqrt3)^2+1))`

Let `tantheta=(x+1)/sqrt3` `=>` `sec^2theta d theta=dx/sqrt3`

Therefore,

`4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/3int(sqrt3 sec^2 theta d theta)/(1+ tan^2theta)`

But `1+tan^2 theta=sec^2theta`

`4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/sqrt3intd theta`

`=4/sqrt3arctan((x+1)/sqrt3)`