# How do I find the integral int1/((w-4)(w+1))dw ?

Jul 28, 2014

$= \frac{1}{5} \ln \left(\frac{w - 4}{w + 1}\right) + c$, where $c$ is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

$\frac{1}{\left(w - 4\right) \left(w + 1\right)}$, it can be written as

$\frac{1}{\left(w - 4\right) \left(w + 1\right)} = \frac{A}{w - 4} + \frac{B}{w + 1}$

multiplying by $\left(w - 4\right) \left(w + 1\right)$ on both sides, we get

$1 = A \left(w + 1\right) + B \left(w - 4\right)$

$1 = \left(A + B\right) w + \left(A - 4 B\right)$

Now comparing coefficient of $w$ and constants both sides, we get

$A + B = 0$ $\implies$ $A = - B$ ...........$\left(i\right)$
$A - 4 B = 1$ ..............$\left(i i\right)$

Substituting value of $A$ from $\left(i\right)$ to $\left(i i\right)$, we get

$- 5 B = 1$ $\implies$ $B = - \frac{1}{5}$

from $B$, we can easily calculate $A$, which will be $\frac{1}{5}$

Now,

$\frac{1}{\left(w - 4\right) \left(w + 1\right)} = \frac{1}{5 \left(w - 4\right)} - \frac{1}{5 \left(w + 1\right)}$

Integrating both side with respect to $w$,

$\int \frac{1}{\left(w - 4\right) \left(w + 1\right)} \mathrm{dw} = \int \frac{1}{5 \left(w - 4\right)} \mathrm{dw} - \int \frac{1}{5 \left(w + 1\right)} \mathrm{dw}$

$= \frac{1}{5} \left(\ln \left(w - 4\right) - \ln \left(w + 1\right)\right) + c$, where $c$ is a constant

$= \frac{1}{5} \ln \left(\frac{w - 4}{w + 1}\right) + c$, where $c$ is a constant