Question

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

Answer 1

The answer is `=-1/(1+s)+C`

Explanation:

You don't need partial fractions

Perform the substitution

`u=1+s`, `du=ds`

`int(ds)/(1+s)^2=int(u)^-2du`

`=-1/(u)`

`=-1/(1+s)+C`

Answer 2

Partial fraction decomposition does not help; it gives you: `1/(s+1)^2`

The integral `int 1/(s+1)^2ds` is best integrated by "u" substitution.

Explanation:

Set up the expansion equation:

`1/(s+1)^2 = A/(s+1)+B/(s+1)^2`

Multiply both sides by `(s+1)^2`:

`1 = A(s+1)+B`

`A = 0, B=1`

The decomposition is the same as the original:

`1/(s+1)^2 = 1/(s+1)^2`

Returning to the integral:

`int 1/(s+1)^2 ds`

let `u = s+1`, then `du = ds` and the integral becomes:

`int u^-2 du = -u^-1 + C`

Reverse the substitution:

`int 1/(s+1)^2 ds = -(s+1)^-1 + C`