Question

How do you integrate `x^4/(x-1)^3` using partial fractions?

Answer 1

The answer is `=-1/(2(x-1)^2)-4/(x-1)+6ln(|x-1|)+x^2/2+3x+C`

Explanation:

The degree of the numerator is greater than the degree of the denominator

Perform a polynomial long division

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

Now, we can perform the decomposition into partial fractions

`(6x^2-8x+3)/(x-1)^3=A/(x-1)^3+B/(x-1)^2+C/(x-1)`

`=(A+B(x-1)+C(x-1)^2)/(x-1)^3`

The denominators are the same, compare the numerators

`(6x^2-8x+3)=A+B(x-1)+C(x-1)^2`

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

Let `x=0`, `=>`, `3=A-B+C`

`=>`, `C-B=2`

Coefficients of `x^2`

`6=C`

`B=6-2=4`

Therefore,

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

So,

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

Finally,

`int(x^4dx)/(x-1)^3=int(dx)/(x-1)^3+4int(dx)/(x-1)^2+6int(dx)/(x-1)+intxdx+int3dx`

`=-1/(2(x-1)^2)-4/(x-1)+6ln(|x-1|)+x^2/2+3x+C`