Calculus I - Substitution Rule for Indefinite Integrals

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Section 5.3 : Substitution Rule for Indefinite Integrals

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17. Evaluate each of the following integrals.

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

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$\displaystyle \int {{\frac{{3x}}{{1 + 9{x^2}}}\,dx}}$ 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,

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

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

$\displaystyle \int {{\frac{3}{{1 + 9{x^2}}}\,dx}}$ 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,

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