Question

How do you integrate `x^2/((x-3)^2(x+4))` using partial fractions?

Answer 1

`intx^2/((x-3)^2(x+4))dx=16/49ln|x+4|+33/49ln|x-3|-9/(7(x-3))+C`

Explanation:

`x^2/((x-3)^2(x+4))=A/(x+4)+B/(x-3)+C/(x-3)^2`
`x^2=A(x-3)^2+B(x-3)(x+4)+C(x+4)`

Substituting `x=3`:
`9=7C` so `C=9/7`

Substituting `x=-4`:
`16=49A` so `A=16/49`

Looking at the coefficients of `x^2` on both sides, we see that `x^2=Ax^2+Bx^2` or that `A+B=1`.
`B=1-16/49=33/49`

Therefore, the partial fraction decomposition of `x^2/((x-3)^2(x+4))` is `(16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2`.

`intx^2/((x-3)^2(x+4))dx=int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dx`

`int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dx=16/49ln|x+4|+33/49ln|x-3|-9/(7(x-3))+C`

Answer 2

`int (x^2dx)/((x-3)^2(x+4)) =33/49ln abs(x-3)-9/(7(x-3)) +16/49ln abs(x+4)+C`

Explanation:

Apply partial fraction decomposition:

`x^2/((x-3)^2(x+4)) = A/(x-3)+B/(x-3)^2 +C/(x+4)`

`x^2/((x-3)^2(x+4)) = (A(x-3)(x+4)+B(x+4)+C(x-3)^2)/((x-3)^2(x+4))`

`x^2 = A(x^2+x-12)+Bx+4B+C(x^2-6x+9))`

`x^2 = Ax^2+ Ax-12A +Bx+4B +Cx^2 -6Cx+9C`

`x^2 = (A+C)x^2 + (A+B-6C)x - (12A-4B-9C)`

`{(A+C=1),(A+B-6C=0),(12A-4B-9C=0):}`

`{(A=1-C),(B-7C=-1),(4B+21C=12):}`

`{(A=1-C),(-4B+28C=4),(4B+21C=12):}`

`{(A=33/49),(B=9/7),(C=16/49):}`

Then:

`int (x^2dx)/((x-3)^2(x+4)) =33/49int dx /(x-3)+9/7int dx/(x-3)^2 +16/49int dx/(x+4)`

`int (x^2dx)/((x-3)^2(x+4)) =33/49ln abs(x-3)-9/(7(x-3)) +16/49ln abs(x+4)+C`