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### Section 7.6 : Integrals Involving Quadratics

3. Evaluate the integral $$\displaystyle \int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}}$$.

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The first thing to do is to complete the square (we’ll leave it to you to verify the completing the square details) on the quadratic in the denominator.

$\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = \int{{\frac{{2t + 9}}{{{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{5}{2}}}}}\,dt}}$ Show Step 2

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

$u = t - 7\hspace{0.5in} \Rightarrow \hspace{0.5in}du = dt\hspace{0.25in}\& \hspace{0.25in}t = u + 7$

Doing the substitution gives,

$\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = \int{{\frac{{2\left( {u + 7} \right) + 9}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}} = \int{{\frac{{2u + 23}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}}$ Show Step 3

Next, we’ll need to split the integral up as follows,

$\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = \int{{\frac{{2u}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}} + \int{{\frac{{23}}{{{{\left( {{u^2} - 3} \right)}^{\frac{5}{2}}}}}\,du}}$

The first integral can be done with the substitution $$v = {u^2} - 3$$ and the second integral will require the trig substitution $$u = \sqrt 3 \sec \theta$$ . Here is the substitution work.

\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23}}{{{{\left( {3{{\sec }^2}\theta - 3} \right)}^{\frac{5}{2}}}}}\left( {\sqrt 3 \sec \theta \tan \theta } \right)\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23\sqrt 3 \sec \theta \tan \theta }}{{{{\left( {3{{\tan }^2}\theta } \right)}^{\frac{5}{2}}}}}\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \int{{\frac{{23\sec \theta }}{{9{{\tan }^4}\theta }}\,d\theta }}\\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{\frac{{{{\cos }^3}\theta }}{{{{\sin }^4}\theta }}\,d\theta }}\end{align*}

Now, for the second integral, don’t forget the manipulations we often need to do so we can do these kinds of integrals. If you need some practice on these kinds of integrals go back to the practice problems for the second section of this chapter and work some of them.

Here is the rest of the integration process for this problem.

\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^4}\theta }}\cos \theta \,d\theta }}\hspace{0.25in}\,w = \sin \theta \\ & = \int{{{v^{ - \,\,\frac{5}{2}}}\,dv}} + \frac{{23}}{9}\int{{{w^{ - 4}} - {w^{ - 2}}\,dw}}\\ & = - \frac{2}{3}{v^{ - \,\,\frac{3}{2}}} + \frac{{23}}{9}\left[ { - \frac{1}{3}{{\left( {\sin \theta } \right)}^{ - 3}} + {{\left( {\sin \theta } \right)}^{ - 1}}} \right] + c\end{align*} Show Step 4

We now need to do quite a bit of back substitution to get the answer back into $$t$$’s. Let’s start with the result of the second integration. Converting the $$\theta$$’s back to $$u$$’s will require a quick right triangle.

 From the substitution we have, $\sec \theta = \frac{u}{{\sqrt 3 }}\,\,\,\,\,\left( { = \frac{{{\mbox{hyp}}}}{{{\mbox{adj}}}}} \right)$ From the right triangle we get, $\sin \theta = \frac{{\sqrt {{u^2} - 3} }}{u}$

Plugging this into the integral above gives,

$\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} = - \frac{2}{{3{{\left( {{u^2} - 3} \right)}^{\,\frac{3}{2}}}}} - \frac{{23{u^3}}}{{27{{\left( {{u^2} - 3} \right)}^{\frac{3}{2}}}}} + \frac{{23u}}{{9\sqrt {{u^2} - 3} }} + c$

Note that we also back substituted for the $$v$$ in the first term as well and rewrote the first term a little. Finally, all we need to do is back substitute for the $$u$$.

\begin{align*}\int{{\frac{{2t + 9}}{{{{\left( {{t^2} - 14t + 46} \right)}^{\frac{5}{2}}}}}\,dt}} & = - \frac{2}{{3{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} - \frac{{23{{\left( {t - 7} \right)}^3}}}{{27{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} + \frac{{23\left( {t - 7} \right)}}{{9\sqrt {{{\left( {t - 7} \right)}^2} - 3} }} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{23\left( {t - 7} \right)}}{{9\sqrt {{{\left( {t - 7} \right)}^2} - 3} }} - \frac{{18 + 23{{\left( {t - 7} \right)}^3}}}{{27{{\left( {{{\left( {t - 7} \right)}^2} - 3} \right)}^{\frac{3}{2}}}}} + c}}\end{align*}

We’ll leave this solution with a final note about these kinds of problems. They are often very long, messy and there are ample opportunities for mistakes so be careful with these and don’t get into too much of a hurry when working them.