Question

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

Answer 1

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:

`1/((x+1)(x+3)(x+5))=A/(x+1)+B/(x+3)+C/(x+5)`

Multiply both sides by the denominator of the LHS.

`=>1/cancel((x+1)(x+3)(x+5))=(A(x+3)(x+5))/cancel(x+1)+(B(x+1)(x+5))/cancel(x+3)+(C(x+1)(x+3))/cancel(x+5)`

`=>1=A(x+3)(x+5)+B(x+1)(x+5)+C(x+1)(x+3)`

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(-2)(2)=-4B =>B=-1/4`

`x=-1:`

`1=A(2)(4)=8A=>A=1/8`

`x=-5:`

`1=C(-4)(-2)=8C=>C=1/8`

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

`=>int(1/8)/(x+1)-(1/4)/(x+3)+(1/8)/(x+5)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:

`1/8ln(|x+1|)-1/4ln(|x+3|)-+1/8ln(|x+5|)+C`

Or, equivalently:

`(ln(|x+1|)-2ln(|x+3|)-+ln(|x+5|))/8+C`