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Q. 49

Expert-verified
Found in: Page 465

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Solve the integral$\int \frac{1}{{x}^{2}\sqrt{{x}^{2}-9}}dx\phantom{\rule{0ex}{0ex}}$

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

See the step by step solution

## Step 1. Given

$\int \frac{1}{{x}^{2}\sqrt{{x}^{2}-9}}dx\phantom{\rule{0ex}{0ex}}$

## Step 2. Using identity

$\mathrm{To}\mathrm{solve}\mathrm{the}\mathrm{integral},\phantom{\rule{0ex}{0ex}}\mathrm{let}\mathrm{x}=3\mathrm{cosh}\left(\mathrm{u}\right).\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}\mathrm{solved}\mathrm{below}.\phantom{\rule{0ex}{0ex}}x=3\mathrm{cos}h\left(u\right)dx\phantom{\rule{0ex}{0ex}}=3\mathrm{sin}h\left(\left(u\right)du\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{use}\mathrm{the}\mathrm{identity}cos{h}^{2}\left(u\right)-1=sin{h}^{2}\left(u\right)\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\frac{1}{{x}^{2}\sqrt{{x}^{2}-9}}=\frac{1}{\left(9\sqrt[]{\left(9\mathrm{cos}{h}^{2}\left(u\right)-9\right)}cos{h}^{2}\left(u\right)\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{27\sqrt{\mathrm{cos}{h}^{2}\left(u\right)}-1cos{h}^{2}\left(u\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{27\sqrt{sin{h}^{2}\left(u}\right)cos{h}^{2}\left(u\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{27sinh\left(u\right)cos{h}^{2}\left(u\right)}$