I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 7.1 : Integration by Parts
4. Evaluate \( \displaystyle \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}}\) .
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The first step here is to pick \(u\) and \(dv\).
Note that if we choose the inverse tangent for \(dv\) the only way to get \(v\) is to integrate \(dv\) and so we would need to know the answer to get the answer and so that won’t work for us. Therefore, the only real choice for the inverse tangent is to let it be \(u\).
So, here are our choices for \(u\) and \(dv\).
\[u = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right)\hspace{0.5in}dv = \,dw\]Don’t forget the \(dw\)! The differential \(dw\) still needs to be put into the \(dv\) even though there is nothing else left in the integral.
Show Step 2Next, we need to compute \(du\) (by differentiating \(u\)) and \(v\) (by integrating \(dv\)).
\[\begin{align*}u & = 6{\tan ^{ - 1}}\left( {\frac{8}{w}} \right) & \hspace{0.5in} & \to & \hspace{0.25in}du & = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + {{\left( {\frac{8}{w}} \right)}^2}}}dw = 6\frac{{ - \frac{8}{{{w^2}}}}}{{1 + \frac{{64}}{{{w^2}}}}}dw\\ dv & = \,dw & \hspace{0.5in} & \to & \hspace{0.25in}v & = w\end{align*}\] Show Step 3In order to complete this problem we’ll need to do some rewrite on \(du\) as follows,
\[du = \frac{{ - 48}}{{{w^2} + 64}}dw\]Plugging \(u\), \(du\), \(v\) and \(dv\) into the Integration by Parts formula gives,
\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = 6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 48\int{{\frac{w}{{{w^2} + 64}}\,dw}}\] Show Step 4Okay, the new integral we get is easily doable (with the substitution \(u = 64 + {w^2}\) ) and so all we need to do to finish this problem out is do the integral.
\[ \int{{6{{\tan }^{ - 1}}\left( {\frac{8}{w}} \right)\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6w\tan ^{ - 1}\left( {\frac{8}{w}} \right) + 24\ln \left| {{w^2} + 64} \right| + c}}\]