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February 18, 2026
Section 5.3 : Substitution Rule for Indefinite Integrals
15. Evaluate \( \displaystyle \int{{\frac{6}{{7 + {y^2}}}\,dy}}\).
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The integrand looks an awful lot like the derivative of the inverse tangent.
\[\frac{d}{{du}}\left( {{{\tan }^{ - 1}}\left( u \right)} \right) = \frac{1}{{1 + {u^2}}}\]So, let’s do a little rewrite to make the integrand look more like this.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \int{{\frac{6}{{7\left( {1 + \frac{1}{7}{y^2}} \right)}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + \frac{1}{7}{y^2}}}\,dy}}\]Let’s do one more rewrite of the integrand.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + {{\left( {\frac{y}{{\sqrt 7 }}} \right)}^2}}}\,dy}}\]At this point we can see that the following substitution will work for us.
\[u = \frac{y}{{\sqrt 7 }}\hspace{0.5in} \to \hspace{0.5in}du = \frac{1}{{\sqrt 7 }}dy\hspace{0.5in} \to \hspace{0.5in}\,dy = \sqrt 7 du\] Show Step 3Doing the substitution and evaluating the integral gives,
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \frac{6}{{\sqrt 7 }}{\tan ^{ - 1}}\left( u \right) + c\]Finally, don’t forget to go back to the original variable!
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{6}{{\sqrt 7 }}{{\tan }^{ - 1}}\left( {\frac{y}{{\sqrt 7 }}} \right) + c}}\]Substitutions for inverse trig functions can be a little tricky to spot when you are first start doing them. Once you do enough of them however they start to become a little easier to spot.