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

Oct 26, 2017

The answer is $= - \frac{1}{1 + s} + C$

#### Explanation:

You don't need partial fractions

Perform the substitution

$u = 1 + s$, $\mathrm{du} = \mathrm{ds}$

$\int \frac{\mathrm{ds}}{1 + s} ^ 2 = \int {\left(u\right)}^{-} 2 \mathrm{du}$

$= - \frac{1}{u}$

$= - \frac{1}{1 + s} + C$

Oct 26, 2017

Partial fraction decomposition does not help; it gives you: $\frac{1}{s + 1} ^ 2$

The integral $\int \frac{1}{s + 1} ^ 2 \mathrm{ds}$ is best integrated by "u" substitution.

#### Explanation:

Set up the expansion equation:

$\frac{1}{s + 1} ^ 2 = \frac{A}{s + 1} + \frac{B}{s + 1} ^ 2$

Multiply both sides by ${\left(s + 1\right)}^{2}$:

$1 = A \left(s + 1\right) + B$

$A = 0 , B = 1$

The decomposition is the same as the original:

$\frac{1}{s + 1} ^ 2 = \frac{1}{s + 1} ^ 2$

Returning to the integral:

$\int \frac{1}{s + 1} ^ 2 \mathrm{ds}$

let $u = s + 1$, then $\mathrm{du} = \mathrm{ds}$ and the integral becomes:

$\int {u}^{-} 2 \mathrm{du} = - {u}^{-} 1 + C$

Reverse the substitution:

$\int \frac{1}{s + 1} ^ 2 \mathrm{ds} = - {\left(s + 1\right)}^{-} 1 + C$