Question

What is the integrate of x^2/(x-1)^3 ?

Answer 1

`int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C`

Explanation:

Part 1

Decompose `x^2/(x-1)^3` into partial fractions

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

Where A, B and C are just numbers

So to solve, just merge those fractions

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

Rearranging

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

By comparing coefficients, you can tell that

`A=1, B-2A=0, A-B+C=0`

Solving, you get `A=1, B=2 and C=1`

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

Part 2

Substituting this back into the integral, you get

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

`int 1/(x-1) dx=log|x-1|+C`

Using the reverse power rule

`int 1/(x-1)^2 dx=int (x-1)^-2 dx=-1/(x-1)+C=1/(1-x)+C`

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

Hence
`int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C`

You can further simplify this to be

`int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C=(3-4x)/(2(x-1)^2)+log|x-1|+C`

P.S. Please classify this question as "Calculus", you are currently marking it as "algebra"

Answer 2

`ln|(x-1)|-2/(x-1)-1/(2(x-1)^2)+C`.

Explanation:

Have a look at this Second Solution :

Let, `I=intx^2/(x-1)^3dx`.

We subst. `x=1+t. :. dx=dt, and, (x-1)=t`.

`:. I=int(t+1)^2/t^3dt=int(t^2+2t+1)/t^3dt`,

`=int{t^2/t^3+2*t/t^3+1/t^3}dt`,

`=int{1/t+2*t^-2+t^-3}dt`,

`=ln|t|+2*t^(-2+1)/(-2+1)+t^(-3+1)/(-3+1)`,

`=ln|t|-2*t^-1-1/2*t^-2`.

`rArr I=ln|(x-1)|-2/(x-1)-1/(2(x-1)^2)+C`, as Respected

e52fa787 has already derived!