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

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

Answered by Sutrisno last updated on 14/Oct/24

$$\int \frac{\sqrt{{\left(3x\right)}^2-7^2}}{x^3}dx$$$$3x=7sec\theta \to dx=\frac{7}{3}sec\theta tan\theta d\theta$$$$=\int \frac{\sqrt{{\left(7sec\theta \right)}^2-7^2}}{{\left(\frac{7}{3}sec\theta \right)}^3}.\frac{7}{3}sec\theta tan\theta d\theta$$$$=\int \frac{7tan\theta }{\frac{343}{27}{sec}^3\theta }.\frac{7}{3}sec\theta tan\theta d\theta$$$$=\frac{9}{7}\int \frac{{tan}^2\theta }{{sec}^2\theta }d\theta$$$$=\frac{9}{7}\int {sin}^2\theta d\theta$$$$=\frac{9}{7}\int \frac{1-cos2\theta }{2}d\theta$$$$=\frac{9}{14}(\theta -\frac{1}{2}sin2\theta )$$$$=\frac{9}{14}(\theta -sin\theta cos\theta )+c$$$$=\frac{9}{14}({sec}^{-1}\left(\frac{3x}{7}\right)-\frac{\sqrt{9x^2-49}}{3x}.\frac{7}{3x})+c$$$$=\frac{9}{14}({sec}^{-1}\left(\frac{3x}{7}\right)-\frac{7\sqrt{9x^2-49}}{9x^2})+c$$

Commented by Hanuda354 last updated on 14/Oct/24

$$\mathrm{Thanks}$$

Answered by Ghisom last updated on 14/Oct/24

$$\int \frac{\sqrt{9x^2-49}}{x^3}dx=$$$$[t=\arcsin \frac{14\sqrt{9x^2-49}}{x^2}\to dx=-\frac{x\sqrt{9x^2-49}}{14}dt]$$$$=-\frac{9}{28}\int (1+\cos t)dt=-\frac{9}{28}(t+\sin t)=$$$$=-\frac{\sqrt{9x^2-49}}{2x^2}-\frac{9}{28}\arcsin \frac{14\sqrt{9x^2-49}}{x^2}+C$$

Commented by Hanuda354 last updated on 14/Oct/24

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

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