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 $\displaystyle \int \frac{dx}{ x \sqrt{x+1} }$.

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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

$$\begin{aligned} \int \frac{2 \, du}{ u^2 - 1 } &=\int \frac{du}{u-1} - \int \frac{du}{u+1} \\ &= \ln \left| u-1 \right| - \ln \left|u+1 \right| + C \\ &=\ln \left| \dfrac{ \sqrt{ x+1 } -1 } { \sqrt{ x + 1 } + 1 }\right| + C, \qquad (\text{since }u=\sqrt{x+1}) \end{aligned}$$

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 $\displaystyle \int \frac{x^4 \, dx}{\sqrt{x^5+5}}.$

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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. $_\square$