Integration of Radical Functions
We will look at various ways to integrate some radical functions using various \(u\)-substitution tricks. These integrals often require making trigonometric substitutions or \(u\)-substitutions to bring them to a simpler form.
Trick: integrals of the form \( \frac{ f'(x) } { f(x) } \)
We've already seen examples of this in Type 2a and Type 3b.\[\]
Trick: integrals involving \( \sqrt{x+a^2} \)
Evaluate \(\displaystyle \int \frac{dx}{ x \sqrt{x+1} } \).
Let \( x + 1 = u^2 \) and \( x = u^2 - 1. \) Then since \( dx = 2u \, du,\) we have
\[ \int \frac{dx}{ x \sqrt{x+1} } = \int \frac{ 2u \, du } { ( u^2 - 1 )\times u } = \int \frac{2 \, du}{ u^2 - 1 }.\]
Then you break up the integrals using partial fractions and obtain
\[\begin{align} \int \frac{2 \, du}{ u^2 - 1 } &=\int \frac{du}{u-1} - \int \frac{du}{u+1} \\ &= \ln \left| u-1 \right| - \ln \left|u+1 \right| + C \\ &=\ln \left| \dfrac{ \sqrt{ x+1 } -1 } { \sqrt{ x + 1 } + 1 }\right| + C, \qquad (\text{since }u=\sqrt{x+1}) \end{align}\]
where \(C\) is the constant of integration. \(_\square\)
Trick: integrals involving \( (x-b)^2 + c^2 \)
We've already seen examples of this in \( \displaystyle \int \frac{x} { \left[ (x-b) ^2 + c^2 \right]^n }\, dx \).
Evaluate \( \displaystyle \int \frac{ 1 } { \sqrt{ x^2 + 2x + 5 } } \, dx \).
Trick: integrals of the form \( \frac{f'} { \sqrt{f} }, \) which integrate to \( 2 \sqrt{f} + C \)
Evaluate \(\displaystyle \int \frac{x^4 \, dx}{\sqrt{x^5+5}}.\)
Multiply and divide by 5, and the integral becomes
\[\frac{1}{5} \int \frac{5x^4 \, dx}{\sqrt{5+x^5}}.\]
Observe that integral is of the form \( \frac{f'} { \sqrt{f} }, \) then the integral solves to
\[2\frac{\sqrt{x^5+5}}{5} + C,\]
where \(C\) is the constant of integration. \(_\square\)