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.

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

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

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

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!