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Section 5.3 : Substitution Rule for Indefinite Integrals

17. Evaluate each of the following integrals.

  1. \( \displaystyle \int {{\frac{{3x}}{{1 + 9{x^2}}}\,dx}}\)
  2. \( \displaystyle \int {{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}}\)
  3. \( \displaystyle \int {{\frac{3}{{1 + 9{x^2}}}\,dx}}\)
Hint : Make sure you pay attention to each of these and note the differences between each integrand and how that will affect the substitution and/or answer.

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a \( \displaystyle \int {{\frac{{3x}}{{1 + 9{x^2}}}\,dx}}\) Show Solution

In this case it looks like the substitution should be

\[u = 1 + 9{x^2}\]

Here is the differential for this substitution.

\[du = 18x\,dx\hspace{0.5in} \Rightarrow \hspace{0.5in}3x\,dx = \frac{1}{6}du\]

The integral is then,

\[\int{{\frac{{3x}}{{1 + 9{x^2}}}\,dx}} = \frac{1}{6}\int{{\frac{1}{u}\,du}} = \frac{1}{6}\ln \left| u \right| + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{6}\ln \left| {1 + 9{x^2}} \right| + c}}\]

b \( \displaystyle \int {{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}}\) Show Solution

The substitution and differential work for this part are identical to the previous part.

\[u = 1 + 9{x^2}\hspace{0.5in}du = 18x\,dx\hspace{0.5in} \Rightarrow \hspace{0.5in}\,3x\,dx = \frac{1}{6}du\]

Here is the integral for this part,

\[\int{{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}} = \frac{1}{6}\int{{\frac{1}{{{u^4}}}\,du}} = \frac{1}{6}\int{{{u^{ - 4}}\,du}} = - \frac{1}{{18}}{u^{ - 3}} + c = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{18}}\frac{1}{{{{\left( {1 + 9{x^2}} \right)}^3}}} + c}}\]

Be careful to not just turn every integral of functions of the form of 1/(something) into logarithms! This is one of the more common mistakes that students often make.

c \( \displaystyle \int {{\frac{3}{{1 + 9{x^2}}}\,dx}}\) Show Solution

Because we no longer have an \(x\) in the numerator this integral is very different from the previous two. Let’s do a quick rewrite of the integrand to make the substitution clearer.

\[\int{{\frac{3}{{1 + 9{x^2}}}\,dx}} = \int{{\frac{3}{{1 + {{\left( {3x} \right)}^2}}}\,dx}}\]

So, this looks like an inverse tangent problem that will need the substitution.

\[u = 3x\hspace{0.5in} \to \hspace{0.5in}du = 3dx\]

The integral is then,

\[\int{{\frac{3}{{1 + 9{x^2}}}\,dx}} = \int{{\frac{1}{{1 + {u^2}}}\,du}} = {\tan ^{ - 1}}\left( u \right) + c = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\tan }^{ - 1}}\left( {3x} \right) + c}}\]