Suggested languages for you:

Americas

Europe

Q. 49

Expert-verifiedFound in: Page 465

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve the integral

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

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

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

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

$\frac{1}{{x}^{2}\sqrt{{x}^{2}-9}}\phantom{\rule{0ex}{0ex}}=\int \frac{1}{9cos{h}^{2}\left(u\right)}du\phantom{\rule{0ex}{0ex}}=\frac{1}{tanh\left(u\right)}+C\phantom{\rule{0ex}{0ex}}=\frac{1}{9tanh\left(cos{h}^{-}1\right(\frac{x}{2}\left)\right)}+C\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{\frac{x}{3}+1}}{3x}\sqrt{\frac{x}{3}-1}+C\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{x-3}}{9x}\sqrt{x+3}+C\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{is}\mathrm{boxed}\frac{\sqrt{x-3}}{9x}\sqrt{x+3}+C$

94% of StudySmarter users get better grades.

Sign up for free