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Question Number 212456 by Hanuda354 last updated on 14/Oct/24

∫ ((√(9x^2 −49))/x^3 ) dx = ?

$$\int\:\:\frac{\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{{x}^{\mathrm{3}} }\:\:{dx}\:=\:? \\ $$

Answered by Sutrisno last updated on 14/Oct/24

∫((√((3x)^2 −7^2 ))/x^3 )dx 3x=7secθ→dx=(7/3)secθtanθdθ =∫((√((7secθ)^2 −7^2 ))/(((7/3)secθ)^3 )).(7/3)secθtanθdθ =∫((7tanθ)/(((343)/(27))sec^3 θ)).(7/3)secθtanθdθ =(9/7)∫((tan^2 θ)/(sec^2 θ))dθ =(9/7)∫sin^2 θdθ =(9/7)∫((1−cos2θ)/2)dθ =(9/(14))(θ−(1/2)sin2θ) =(9/(14))(θ−sinθcosθ)+c =(9/(14))(sec^(−1) (((3x)/7))−((√(9x^2 −49))/(3x)).(7/(3x)))+c =(9/(14))(sec^(−1) (((3x)/7))−((7(√(9x^2 −49)))/(9x^2 )))+c

$$\int\frac{\sqrt{\left(\mathrm{3}{x}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }}{{x}^{\mathrm{3}} }{dx} \\ $$$$\mathrm{3}{x}=\mathrm{7}{sec}\theta\rightarrow{dx}=\frac{\mathrm{7}}{\mathrm{3}}{sec}\theta{tan}\theta{d}\theta \\ $$$$=\int\frac{\sqrt{\left(\mathrm{7}{sec}\theta\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }}{\left(\frac{\mathrm{7}}{\mathrm{3}}{sec}\theta\right)^{\mathrm{3}} }.\frac{\mathrm{7}}{\mathrm{3}}{sec}\theta{tan}\theta{d}\theta \\ $$$$=\int\frac{\mathrm{7}{tan}\theta}{\frac{\mathrm{343}}{\mathrm{27}}{sec}^{\mathrm{3}} \theta}.\frac{\mathrm{7}}{\mathrm{3}}{sec}\theta{tan}\theta{d}\theta \\ $$$$=\frac{\mathrm{9}}{\mathrm{7}}\int\frac{{tan}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\frac{\mathrm{9}}{\mathrm{7}}\int{sin}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\mathrm{9}}{\mathrm{7}}\int\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\mathrm{9}}{\mathrm{14}}\left(\theta−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta\right) \\ $$$$=\frac{\mathrm{9}}{\mathrm{14}}\left(\theta−{sin}\theta{cos}\theta\right)+{c} \\ $$$$=\frac{\mathrm{9}}{\mathrm{14}}\left({sec}^{−\mathrm{1}} \left(\frac{\mathrm{3}{x}}{\mathrm{7}}\right)−\frac{\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{3}{x}}.\frac{\mathrm{7}}{\mathrm{3}{x}}\right)+{c} \\ $$$$=\frac{\mathrm{9}}{\mathrm{14}}\left({sec}^{−\mathrm{1}} \left(\frac{\mathrm{3}{x}}{\mathrm{7}}\right)−\frac{\mathrm{7}\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{9}{x}^{\mathrm{2}} }\right)+{c} \\ $$

Commented by Hanuda354 last updated on 14/Oct/24

Thanks

$$\mathrm{Thanks} \\ $$

Answered by Ghisom last updated on 14/Oct/24

∫ ((√(9x^2 −49))/x^3 )dx= [t=arcsin ((14(√(9x^2 −49)))/x^2 ) → dx=−((x(√(9x^2 −49)))/(14))dt] =−(9/(28))∫(1+cos t)dt=−(9/(28))(t+sin t)= =−((√(9x^2 −49))/(2x^2 ))−(9/(28))arcsin ((14(√(9x^2 −49)))/x^2 ) +C

$$\int\:\frac{\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{{x}^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\frac{\mathrm{14}\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{{x}^{\mathrm{2}} }\:\rightarrow\:{dx}=−\frac{{x}\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{14}}{dt}\right] \\ $$$$=−\frac{\mathrm{9}}{\mathrm{28}}\int\left(\mathrm{1}+\mathrm{cos}\:{t}\right){dt}=−\frac{\mathrm{9}}{\mathrm{28}}\left({t}+\mathrm{sin}\:{t}\right)= \\ $$$$=−\frac{\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{\mathrm{2}{x}^{\mathrm{2}} }−\frac{\mathrm{9}}{\mathrm{28}}\mathrm{arcsin}\:\frac{\mathrm{14}\sqrt{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{49}}}{{x}^{\mathrm{2}} }\:+{C} \\ $$

Commented by Hanuda354 last updated on 14/Oct/24

Thank you

$$\mathrm{Thank}\:\mathrm{you} \\ $$

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