Question

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{1}{{\left(x^2+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx$

Short answer

The solution of the integral is$\frac{x}{4\sqrt{x^2+4}}+C.$

Step-by-step solution

Step 1. Given Information.

The given integral is$\int \frac{1}{{\left(x^2+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx.$

Step 2. Solve. 

To solve the integral, let $x=2\sin u,$ so derivation of uis $dx=2\cos udu.$

Thus, substitute u into the original integral,

role="math" localid="1648802019739" $$\begin{gathered} \int \frac{1}{{\left(x^2+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx \\ =\int \frac{1}{{\left({\left(2\sin u\right)}^2+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}2\cos udu \\ =\int \frac{1}{{\left(4{\sin }^{2}u+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}2\cos udu \\ =\int \frac{1}{{\left(4({\sin }^{2}u+1)\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}2\cos udu \\ =\int \frac{1}{8{\left({\sin }^{2}u+4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}2\cos udu \\ =\frac{1}{4}\int \frac{1}{{\left({\sin }^{2}u+1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\cos udu \\ \mathrm{Let}\text{'}\mathrm{s}\mathrm{use}\mathrm{the}\mathrm{identity}si{n}^{2}x+co{s}^{2}x=1 \\ =\frac{1}{4}\int \frac{1}{{\left({\cos }^{2}u\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\cos udu \\ =\frac{1}{4}\int \frac{1}{{\cos }^{2}u}du \\ =\frac{1}{4}\tan udu \end{gathered}$$

Step 3. Solve. 

By proceeding with the calculation further, substitute back uin the above equation,

$$\begin{gathered} =\frac{1}{4}\tan \left({\sin }^{-1}\left(\frac{x}{2}\right)\right)du \\ =\frac{x}{4\sqrt{x^2+4}}+C \end{gathered}$$