Question

We can extend the technique of trigonometric substitution to the hyperbolic functions. Use Theorem 2.20 and the identity ${\cosh }^{2}x-{\sinh }^{2}x=1$to solve each integral in Exercises 87– 90 with an appropriate hyperbolic substitution $x=a\sinh u\mathrm{or}x=a\cos hu$. (These exercises involve hyperbolic functions.)

$\int \frac{x^2}{{\left(x^2+1\right)}^{3/2}}dx$

Short answer

The value is$\frac{1}{2}\ln \left(\left|e^{2{\sinh }^{-1}\left(x\right)}\right|\right)+\frac{2}{e^{2{\sinh }^{-1}\left(x\right)}+1}+C.$

Step-by-step solution

Step 1. Given Information.

The given integral is$\int \frac{x^2}{{\left(x^2+1\right)}^{3/2}}dx$.

Step 2. Substitution.

Let$x=\sinh \left(u\right)$

Therefore,

$$\begin{gathered} dx=\cosh \left(u\right)du \\ Weknow, \\ {\sinh }^2\left(u\right)+1={\cosh }^2\left(u\right) \\ \frac{x^2}{{\left(x^2+1\right)}^{\frac{3}{2}}}=\frac{{\sinh }^2\left(u\right)}{{\left({\sinh }^2\left(u\right)+1\right)}^{\frac{3}{2}}} \\ =\frac{{\sinh }^2\left(u\right)}{{\left({\cosh }^2\left(u\right)\right)}^{\frac{3}{2}}} \\ =\frac{{\sinh }^2\left(u\right)}{{\cosh }^3\left(u\right)} \end{gathered}$$

Step 3. Value of the integral.

Now, we can write,

$$\begin{gathered} \int \frac{{\sinh }^2\left(u\right)}{{\cosh }^2\left(u\right)}dx=\int \tanh \left(u\right)du \\ =\int \frac{{\left(e^u-e^{-u}\right)}^2}{{\left(e^u+e^{-u}\right)}^2}du \\ =\frac{1}{2}\ln \left(\left|e^{2{\sinh }^{-1}\left(x\right)}\right|\right)+\frac{2}{e^{2{\sinh }^{-1}\left(x\right)}+1}+C \end{gathered}$$