From this we can see that the following substitution should work for us.

\[u = x - 1\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dx\hspace{0.25in}\& \hspace{0.25in}x = u + 1\]

Doing the substitution gives,

\[\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} = \int{{\frac{{10\left( {u + 1} \right)}}{{4{u^2} + 5}}\,du}}\] Show Step 3

We can quickly do this integral if we split it up as follows,

\[\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} = \int{{\frac{{10u}}{{4{u^2} + 5}}\,du}} + \int{{\frac{{10}}{{4{u^2} + 5}}\,du}} = \int{{\frac{{10u}}{{4{u^2} + 5}}\,du}} + \frac{5}{2}\int{{\frac{1}{{{u^2} + \frac{5}{4}}}\,du}}\]

After a quick rewrite of the second integral we can see that we can do the first with the substitution \(v = 4{u^2} + 5\) and the second is an inverse trig integral we can evaluate using the formula given at the start of the notes for this section.

\[\begin{align*}\int{{\frac{{10x}}{{4{x^2} - 8x + 9}}\,dx}} &= \frac{5}{4}\ln \left| v \right| + \frac{5}{2}\left( {\frac{2}{{\sqrt 5 }}} \right){\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 5 }}} \right) + c\\ & = \frac{5}{4}\ln \left| {4{u^2} + 5} \right| + \sqrt 5 {\tan ^{ - 1}}\left( {\frac{{2u}}{{\sqrt 5 }}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}\ln \left| {4{{\left( {x - 1} \right)}^2} + 5} \right| + \sqrt 5 {{\tan }^{ - 1}}\left( {\frac{{2x - 2}}{{\sqrt 5 }}} \right) + c}}\end{align*}\]

Don’t forget to back substitute in for \(u\)!