Question

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

Answer 1

`intx^3/(x^2+2x+1)dx=x^2-2x+3ln|x+1|+1/(x+1)`

Explanation:

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

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

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

i.e. `3x+2=A(x+1)+B`

Now if `x=0`, we have `A+B=2` and if `x=-1`, we have `B=-1` and hence `A=3`

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

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

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

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

Answer 2

`(x+1)^2/2-3(x+1)+3ln|x+1|+1/(x+1)+C.`

Explanation:

Though the Problem is requred to be solved using Partial Fractions,

as a Second Method, we show that the same can be solved

easily without its use.

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

Sbst.ing, `x+1=t," so that, "dx=dt, and, x=t-1.`

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

`=int{t^3-1-3t(t-1)}/t^2dt,`

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

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

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

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

`rArr I=(x+1)^2/2-3(x+1)+3ln|x+1|+1/(x+1)+C.`

Enjoy Maths.!

N.B.: `I=x^2/2+x+1/2-3x-3+3ln|x+1|+1/(x+1)+C,`

`=x^2/2-2x-5/2+3ln|x+1|+1/(x+1)+C,`

`=x^2/2-2x+3ln|x+1|+1/(x+1)+c, where, c=C-5/2,` as

Respected Shwetank Mauria has derived!