# How do you integrate int(x+2)/(x+1)^3 dx =?

Apr 8, 2018

$I = - \left(\frac{1}{x + 1} + \frac{1}{2 {\left(x + 1\right)}^{2}}\right) + C$

#### Explanation:

We want to integrate

$I = \int \frac{x + 2}{x + 1} ^ 3 \mathrm{dx}$

Rewrite the integrand

$I = \int \frac{x + 1 + 1}{x + 1} ^ 3 \mathrm{dx}$

$\textcolor{w h i t e}{I} = \int \frac{x + 1}{x + 1} ^ 3 + \frac{1}{x + 1} ^ 3 \mathrm{dx}$

$\textcolor{w h i t e}{I} = \int \frac{1}{x + 1} ^ 2 + \frac{1}{x + 1} ^ 3 \mathrm{dx}$

By the power rule for integration

$I = - \frac{1}{x + 1} - \frac{1}{2 {\left(x + 1\right)}^{2}} + C$

$\textcolor{w h i t e}{I} = - \left(\frac{1}{x + 1} + \frac{1}{2 {\left(x + 1\right)}^{2}}\right) + C$

Apr 8, 2018

This integral can be solved by either separating the term by the use of partial fraction decomposition then integration or just by, sneakily, splitting the fraction into two terms and integrating.

#### Explanation:

The first method using partial fractions begins by allowing the inner expression to be equal to the following terms:
$\frac{x + 2}{x + 1} ^ 3 = \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1} ^ 3$

(Note: this partial fraction decomposition is a special case when a term in the denominator has a degree greater than 1)

The fractions can be multiplied by ${\left(x + 1\right)}^{3}$ to simplify the equation into a polynomial where then the coefficients of the different terms are equated on the left and right hand sides of the equation:
$x + 2 = A {\left(x + 1\right)}^{2} + B \left(x + 1\right) + C$
$x + 2 = A {x}^{2} + 2 A x + A + B x + B + C$

Hence:
$0 = A$
$1 = 2 A + B$
$2 = A + B + C$

To find the coefficient the unknowns, they can be solved as simultaneous equations yielding:
$A = 0$
$B = 1$
$C = 1$

This means that the fraction can be decomposed to:
$\frac{x + 2}{x + 1} ^ 3 = \frac{1}{x + 1} ^ 2 + \frac{1}{x + 1} ^ 3$

Now the integration should be straightforward.
After the partial fraction decomposition the integral can be rewritten as:
$\int \left(\frac{1}{x + 1} ^ 2 + \frac{1}{x + 1} ^ 3\right) \mathrm{dx}$

$\int \frac{x + 2}{x + 1} ^ 3 \mathrm{dx} = - \frac{1}{x + 1} - \frac{1}{2 {\left(x + 1\right)}^{2}} + C$
$\int \frac{x + 2}{x + 1} ^ 3 \mathrm{dx} = \int \frac{x + 1}{x + 1} ^ 3 + \frac{1}{x + 1} ^ 3 \mathrm{dx}$
$= \int \left(\frac{1}{x + 1} ^ 2 + \frac{1}{x + 1} ^ 3\right) \mathrm{dx}$