Question

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

(a) with the substitution $u=x^2+4x;$

(b) by completing the square and then applying the trigonometric substitution x + 2 = 2 sec u.

Short answer

Part (a) The solution of the integral is $-\frac{1}{\sqrt{{\mathrm{x}}^2-4\mathrm{x}}}+\mathrm{C}.$

Part (b) The solution of the integral is $-\frac{1}{\sqrt{{\mathrm{x}}^2-4\mathrm{x}}}+\mathrm{C}.$

Step-by-step solution

Part (a) Step 1. Given Information.

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

Part (a) Step 2. Solve. 

We have to solve the integral with the substitution $u=x^2+4x,$so the derivative is$du=\left(2x+4\right)dx.$

Let's solve the integral by substituting u,

$$\begin{gathered} \int \frac{x+2}{{\left(x^2+4x\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx \\ =\int \frac{x+2}{{\left(u\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{du}{\left(2x+4\right)} \\ =\frac{1}{2}\int \frac{du}{{\left(u\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ =\frac{1}{2}\left[-2\frac{1}{\sqrt{u}}\right]+C \\ =-\frac{1}{\sqrt{u}}+C \\ \mathrm{Substitute}\mathrm{back}u, \\ =-\frac{1}{\sqrt{x^2+4x}}+C \end{gathered}$$

Part (b) Step 1. Solve.

We have to solve it first by completing the square.

So, let's complete the square,

$$\begin{gathered} x^2+4x \\ =x^2+4x+{\left(2\right)}^2-{\left(2\right)}^2 \\ ={\left(x+2\right)}^2-4 \end{gathered}$$

Part (b) Step 2. Solve.

Now, we have to apply the trigonometric substitution $x+2=2secu.$

So, Let's solve the integral by substituting,

$$\begin{gathered} \int \frac{x+2}{{\left(x^2+4x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx \\ =\int \frac{x+2}{{\left({\left(x+2\right)}^2-4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx \\ =\int \frac{2secu}{{\left({\left(2secu\right)}^2-4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\left(2secu\tan u\right)du \\ =\int \frac{4se{c}^{2}u\tan u}{{\left(4se{c}^{2}u-4\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}du \\ =\int \frac{4se{c}^{2}u\tan u}{8{\left(se{c}^{2}u-1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}du \\ =\frac{1}{2}\int \frac{se{c}^{2}u\tan u}{{\left(se{c}^{2}u-1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}du \\ \mathrm{Use}\mathrm{the}\mathrm{identity}1+{\tan }^{2}x=se{c}^{2}x, \\ =\frac{1}{2}\int \frac{se{c}^{2}u\tan u}{{\left({\tan }^{2}u\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}du \\ =\frac{1}{2}\int \frac{se{c}^{2}u}{{\tan }^{2}u}du \\ =\frac{1}{2}\int cs{c}^{2}udu \\ =\frac{1}{2}\left(-cotu\right)+C \\ \mathrm{Substitute}\mathrm{back}\mathrm{u}, \\ =-\frac{1}{2}\left(\cot \left({\sec }^{-1}\left(\frac{\mathrm{x}+2}{2}\right)\right)\right)+\mathrm{C} \\ =-\frac{1}{\sqrt{{\mathrm{x}}^2-4\mathrm{x}}}+\mathrm{C} \end{gathered}$$