Now, the first integral looks like it might be an inverse tangent (although we’ll need to do a rewrite of that integral) and the second looks like it’s a logarithm (with a quick substitution).

So, here is the rewrite on the first integral.

\[\int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}} = \frac{8}{9}\int{{\frac{1}{{\frac{4}{9}{w^2} + 1}}\,dw}} - \int{{\frac{w}{{4{w^2} + 9}}\,dw}}\] Show Step 3

Now we’ll need a substitution for each integral. Here are the substitutions we’ll need for each integral.

\[u = \frac{2}{3}w\,\,\,\,\,\,\left( {{\mbox{so }}{u^2} = \frac{4}{9}{w^2}} \right)\hspace{0.75in}v = 4{w^2} + 9\] Show Step 4

Here is the differential work for the substitution.

\[du = \frac{2}{3}dw\hspace{0.25in} \to \hspace{0.25in}dw = \frac{3}{2}du\hspace{0.5in}dv = 8w\,dw\,\,\,\,\,\,\, \to \hspace{0.25in}\,w\,dw = \frac{1}{8}dv\]

Now, doing the substitutions and evaluating the integrals gives,

\[\begin{align*}\int{{\frac{{8 - w}}{{4{w^2} + 9}}\,dw}} & = \frac{8}{9}\left( {\frac{3}{2}} \right)\int{{\frac{1}{{{u^2} + 1}}\,du}} - \frac{1}{8}\int{{\frac{1}{v}\,dv}} = \frac{4}{3}{\tan ^{ - 1}}\left( u \right) - \frac{1}{8}\ln \left| v \right| + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{4}{3}{{\tan }^{ - 1}}\left( {\frac{2}{3}w} \right) - \frac{1}{8}\ln \left| {4{w^2} + 9} \right| + c}}\end{align*}\]

Do not forget to go back to the original variable after evaluating the integral!