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

Feb 8, 2016

$\int \frac{1}{x {\left(x + 1\right)}^{2}} \mathrm{dx} = \ln | x | - \ln | x + 1 | + \frac{1}{x + 1} + \text{Constant}$

#### Explanation:

As stated by the question, we must first express $\frac{1}{x {\left(x + 1\right)}^{2}}$ in partial fractions, before we perform the integration.

$\frac{1}{x {\left(x + 1\right)}^{2}} \equiv \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$,

where $A$, $B$ and $C$ are constants to be determined, such that the equality holds true for all possible values of $x$.

Which means,

$1 \equiv A {\left(x + 1\right)}^{2} + B x \left(x + 1\right) + C x$.

Simplifying gives

$\left(A + B\right) \cdot {x}^{2} + \left(2 A + B + C\right) \cdot x + A \equiv 1$

We compare the coefficients (to zero) to get 3 simultaneous linear equations,

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

The system of equations is solved easily to yield

$A = 1$
$B = - 1$
$C = - 1$

Therefore,

$\frac{1}{x {\left(x + 1\right)}^{2}} \equiv \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2$.

Now, to integrate.

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

$= \ln | x | - \ln | x + 1 | + \frac{1}{x + 1} + \text{Constant}$