Setting the numerators equal gives,

\[8 = A\left( {3x + 4} \right)\left( {x + 1} \right) + Bx\left( {x + 1} \right) + C\,x\left( {3x + 4} \right)\] Show Step 3

We can use the “trick” discussed in the notes to easily get the coefficients in this case so let’s do that. Here is that work.

\[\begin{align*}x = & - \frac{4}{3}: & 8 = & \frac{4}{9}B \\ & & & \\x = & - 1 \,\, : & 8 = & - C\\ & & & \\ x = & 0 \,\,\,\,\,\,\,\,\,: & 8 = & 4A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = 2}\\ & \\ & {B = 18}\\ & \\ & {C = - 8}\end{aligned}\]

The partial fraction form of the integrand is then,

\[\frac{8}{{x\left( {3x + 4} \right)\left( {x + 1} \right)}} = \frac{2}{x} + \frac{{18}}{{3x + 4}} - \frac{8}{{x + 1}}\] Show Step 4

We can now do the integral.

\[\int{{\frac{8}{{x\left( {3x + 4} \right)\left( {x + 1} \right)}}\,dx}} = \int{{\frac{2}{x} + \frac{{18}}{{3x + 4}} - \frac{8}{{x + 1}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{2\ln \left| x \right| + 6\ln \left| {3x + 4} \right| - 8\ln \left| {x + 1} \right| + c}}\]

Hopefully you are getting good enough with integration that you can do some of these integrals in your head. Be careful however with the second integral. When doing these kinds of integrals in our head it is easy to forget about the substitutions that are technically required to do them and then miss the coefficients from the substitutions that need to show up in the answer.