Question

How do you integrate `int (x^2-2x+1)/(x-2)^3` using partial fractions?

Answer 1

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

Explanation:

We can solve the Problem without decomposing the

Integrand into Partial Fractions as shown below :

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

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

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

`=int[{(x-2)+2}/(x-2)^2+1/(x-2)^3]dx`

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

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

Now, we use the following Result :

`intf(x)dx=F(x)+crArrintf(ax+b)=1/aF(ax+b),ane0`.

[ It can be easily proved by subst.ing, for `ax+b`.]

`:. I=ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.`

Enjoy Maths. , and, spread the Joy!

Answer 2

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

Explanation:

Noticing that `(x-2)^3` is a perfect cube, there is no need for partial fractions. Substitute `u=x-2` and `(du)/(dx)=1` and the integral becomes:
`int((u+2)^2-2(u+2)+1)/u^3 xx 1 du`
`=int(u^2+2u+1)/u^3du`
`=int1/u+2/u^2+1/u^3du`
`= ln|x-2|-2/(x-2)-1/{2(x-2)^2}+C.`