# How do you evaluate the integral int dx/((x+1)(x+3)(x+5))?

Mar 6, 2017

There are a variety of possible methods, but that which sticks out to me is the method of partial fractions. The denominator is already factored for you, so we have:

$\frac{1}{\left(x + 1\right) \left(x + 3\right) \left(x + 5\right)} = \frac{A}{x + 1} + \frac{B}{x + 3} + \frac{C}{x + 5}$

Multiply both sides by the denominator of the LHS.

$\implies \frac{1}{\cancel{\left(x + 1\right) \left(x + 3\right) \left(x + 5\right)}} = \frac{A \left(x + 3\right) \left(x + 5\right)}{\cancel{x + 1}} + \frac{B \left(x + 1\right) \left(x + 5\right)}{\cancel{x + 3}} + \frac{C \left(x + 1\right) \left(x + 3\right)}{\cancel{x + 5}}$

$\implies 1 = A \left(x + 3\right) \left(x + 5\right) + B \left(x + 1\right) \left(x + 5\right) + C \left(x + 1\right) \left(x + 3\right)$

Now we pick values of $x$ so that we can solve for A, B, and C. Alternatively, you could solve the equations simultaneously.

I would use $x = - 3 , - 1 ,$ and $- 5$. This will cancel two of the variables at a time and allow us to solve for that which remains in each case.

$x = - 3 :$

$1 = B \left(- 2\right) \left(2\right) = - 4 B \implies B = - \frac{1}{4}$

$x = - 1 :$

$1 = A \left(2\right) \left(4\right) = 8 A \implies A = \frac{1}{8}$

$x = - 5 :$

$1 = C \left(- 4\right) \left(- 2\right) = 8 C \implies C = \frac{1}{8}$

Now we replace A, B, and C in the above partial fractions and substitute into the integral.

$\implies \int \frac{\frac{1}{8}}{x + 1} - \frac{\frac{1}{4}}{x + 3} + \frac{\frac{1}{8}}{x + 5} \mathrm{dx}$

We can split this into three separate integrals and use a substitution to solve, but it is easily done mentally to give the final answer:

$\frac{1}{8} \ln \left(| x + 1 |\right) - \frac{1}{4} \ln \left(| x + 3 |\right) - + \frac{1}{8} \ln \left(| x + 5 |\right) + C$

Or, equivalently:

$\frac{\ln \left(| x + 1 |\right) - 2 \ln \left(| x + 3 |\right) - + \ln \left(| x + 5 |\right)}{8} + C$