Question Number 153535 by gsk2684 last updated on 08/Sep/21
$${find}\:\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}^{\mathrm{3}} }\:{dx}=? \\ $$
Answered by puissant last updated on 08/Sep/21
![K=∫((√(x^2 −9))/x^3 )dx x=3sec(u) → dx=3sec(u)tan(u)du we have sec(u)=(x/3) ⇒ cos(u)=(3/x) and sin(u)=((√(x^2 −9))/x) K=∫((√(9sec^2 u−9))/(27sec^3 u))×3sec(u)tan(u)du =∫((3tan(u))/(27sec^2 u))×3tan(u)du =(1/3)∫tan^2 u cos^2 u du =(1/6)∫2sin^2 u du = (1/6)∫(1−cos2u)du =(1/6)[u−((sin2u)/2)]+C =(1/6)[u−sin(u)cos(u)]+C ⇒ K=(1/6)[arccos((3/x))−(3/x)×((√(x^2 −9))/x)]+C ∴∵ K=(1/6)arccos((3/x))−((√(x^2 −9))/(2x^2 ))+C](https://www.tinkutara.com/question/Q153546.png)
$${K}=\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}^{\mathrm{3}} }{dx} \\ $$$${x}=\mathrm{3}{sec}\left({u}\right)\:\rightarrow\:{dx}=\mathrm{3}{sec}\left({u}\right){tan}\left({u}\right){du} \\ $$$${we}\:{have}\:{sec}\left({u}\right)=\frac{{x}}{\mathrm{3}}\:\Rightarrow\:{cos}\left({u}\right)=\frac{\mathrm{3}}{{x}} \\ $$$${and}\:{sin}\left({u}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}} \\ $$$${K}=\int\frac{\sqrt{\mathrm{9}{sec}^{\mathrm{2}} {u}−\mathrm{9}}}{\mathrm{27}{sec}^{\mathrm{3}} {u}}×\mathrm{3}{sec}\left({u}\right){tan}\left({u}\right){du} \\ $$$$=\int\frac{\mathrm{3}{tan}\left({u}\right)}{\mathrm{27}{sec}^{\mathrm{2}} {u}}×\mathrm{3}{tan}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int{tan}^{\mathrm{2}} {u}\:{cos}^{\mathrm{2}} {u}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{2}{sin}^{\mathrm{2}} {u}\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\int\left(\mathrm{1}−{cos}\mathrm{2}{u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{u}−\frac{{sin}\mathrm{2}{u}}{\mathrm{2}}\right]+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{u}−{sin}\left({u}\right){cos}\left({u}\right)\right]+{C} \\ $$$$\Rightarrow\:{K}=\frac{\mathrm{1}}{\mathrm{6}}\left[{arccos}\left(\frac{\mathrm{3}}{{x}}\right)−\frac{\mathrm{3}}{{x}}×\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}}\right]+{C} \\ $$$$ \\ $$$$\therefore\because\:\:{K}=\frac{\mathrm{1}}{\mathrm{6}}{arccos}\left(\frac{\mathrm{3}}{{x}}\right)−\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{\mathrm{2}{x}^{\mathrm{2}} }+{C} \\ $$
Commented by gsk2684 last updated on 13/Sep/21
$${thank}\:{you} \\ $$