Question Number 179405 by Acem last updated on 29/Oct/22
$$\:{Evaluate}\:\int\:\frac{{dx}}{{x}^{\mathrm{4}} \:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$
Answered by greougoury555 last updated on 29/Oct/22
$$\:{M}=\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:{x}^{\mathrm{2}} \:\sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx} \\ $$$$\:{M}=\:\int\:\frac{{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\:{let}\:{j}^{\mathrm{2}} \:=\:\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}^{−\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\mathrm{2}{j}\:{dj}\:=−\mathrm{18}{x}^{−\mathrm{3}} \:{dx} \\ $$$$\:{M}=\:\int\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{9}}{j}\right){dj}}{\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right).{j}}=−\int\:\frac{{dj}}{{j}^{\mathrm{2}} +\mathrm{1}} \\ $$$${M}=−\:\mathrm{arctan}\:\left(\sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\:−\mathrm{1}}\:\right)+{c} \\ $$$$\:{M}=\:−\:\mathrm{arctan}\:\left(\frac{{x}}{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}\:\right)+\:{c}\: \\ $$$$ \\ $$
Commented by Acem last updated on 29/Oct/22
$$\:{M}=\:\int\:\frac{{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\:{M}=\:\int\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{9}}{j}\right){dj}}{\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right).{j}}=−\int\:\frac{{dj}}{{j}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}^{\mathrm{2}} \:\neq\:\frac{\mathrm{1}}{\mathrm{9}}\:\left({j}^{\mathrm{2}} \:+\:\mathrm{1}\right) \\ $$
Commented by greougoury555 last updated on 29/Oct/22