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Section 5.4 : More Substitution Rule

12. Evaluate \( \displaystyle \int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}}\).

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Hint : With the integrand written as it is here this problem can’t be done.
Start Solution

As written we can’t do this problem. In order to do this integral we’ll need to rewrite the integral as follows.

\[\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} = \int{{\frac{{7x}}{{\sqrt {1 - 25{x^2}} }}\,dx}} + \int{{\frac{2}{{\sqrt {1 - 25{x^2}} }}\,dx}}\] Show Step 2

Now, the second integral looks like it might be an inverse sine (although we’ll need to do a rewrite of that integral) and the first looks like a simple substitution will work for us.

So, here is the rewrite on the second integral.

\[\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} = \int{{\frac{{7x}}{{\sqrt {1 - 25{x^2}} }}\,dx}} + 2\int{{\frac{1}{{\sqrt {1 - {{\left( {5x} \right)}^2}} }}\,dx}}\] Show Step 3

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

\[u = 1 - 25{x^2}\hspace{0.5in}v = 5x\,\,\,\,\,\,\left( {{\mbox{so }}{v^2} = 25{x^2}} \right)\] Show Step 4

Here is the differential work for the substitution.

\[du = - 50x\,dx\hspace{0.25in} \to \hspace{0.25in}x\,dx = - \frac{1}{{50}}du\hspace{0.5in}dv = 5dx\hspace{0.25in} \to \hspace{0.25in}dx = \frac{1}{5}dv\]

Now, doing the substitutions and evaluating the integrals gives,

\[\begin{align*}\int{{\frac{{7x + 2}}{{\sqrt {1 - 25{x^2}} }}\,dx}} & = - \frac{7}{{50}}\int{{{u^{ - \,\frac{1}{2}}}\,du}} + \frac{2}{5}\int{{\frac{1}{{\sqrt {1 - {v^2}} }}\,dv}} = - \frac{7}{{25}}{u^{\frac{1}{2}}} + \frac{2}{5}{\sin ^{ - 1}}\left( v \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{7}{{25}}{{\left( {1 - 25{x^2}} \right)}^{\frac{1}{2}}} + \frac{2}{5}{{\sin }^{ - 1}}\left( {5x} \right) + c}}\end{align*}\]

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