Question

Solve the integral

$\int \frac{1}{x^2\sqrt{x^2-9}}dx$

Short answer

$\frac{\sqrt{x-3}}{9x}\sqrt{x+3}+C$

Step-by-step solution

Step 1. Given

$\int \frac{1}{x^2\sqrt{x^2-9}}dx$

Step 2. Using identity

$$\begin{gathered} \mathrm{To}\mathrm{solve}\mathrm{the}\mathrm{integral}, \\ \mathrm{let}\mathrm{x}=3\cosh \left(\mathrm{u}\right). \\ \mathrm{Now},\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}\mathrm{solved}\mathrm{below}. \\ x=3\cos h\left(u\right)dx \\ =3\sin h\left(\right(u)du) \\ \mathrm{Now},\mathrm{use}\mathrm{the}\mathrm{identity}cos{h}^2\left(u\right)-1=sin{h}^2\left(u\right) \\ So, \\ \frac{1}{x^2\sqrt{x^2-9}}=\frac{1}{\left(9\sqrt[]{\left(9\cos h^2\right(u)-9)}cos{h}^2\right(u\left)\right)} \\ =\frac{1}{27\sqrt{\cos h^2\left(u\right)}-1cos{h}^2\left(u\right)} \\ =\frac{1}{27\sqrt{sin{h}^2(u}\left)cos{h}^2\right(u)} \\ =\frac{1}{27sinh\left(u\right)cos{h}^2\left(u\right)} \end{gathered}$$

Step 3. Calculation

$$\begin{gathered} \frac{1}{x^2\sqrt{x^2-9}} \\ =\int \frac{1}{9cos{h}^2\left(u\right)}du \\ =\frac{1}{tanh\left(u\right)}+C \\ =\frac{1}{9tanh\left(cos{h}^{-}1\right(\frac{x}{2}\left)\right)}+C \\ =\frac{\sqrt{{\displaystyle \frac{x}{3}}+1}}{3x}\sqrt{\frac{x}{3}-1}+C \\ =\frac{\sqrt{x-3}}{9x}\sqrt{x+3}+C \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{is}\mathrm{boxed}\frac{\sqrt{x-3}}{9x}\sqrt{x+3}+C \end{gathered}$$