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Question Number 122098 by Rohit412 last updated on 14/Nov/20

Commented by liberty last updated on 14/Nov/20

$$\int \frac{\mathrm{dx}}{\sqrt{{(2\mathrm{ax}+{\mathrm{x}}^2)}^3}}=\int \frac{\mathrm{dx}}{{\left(\sqrt{2\mathrm{ax}+{\mathrm{x}}^2}\right)}^3}$$$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^3{\left(\sqrt{{2\mathrm{ax}}^{-1}+1}\right)}^3}=\int \frac{{\mathrm{x}}^{-2}\mathrm{dx}}{\mathrm{x}\sqrt{{({2\mathrm{ax}}^{-1}+1)}^3}}$$$$\mathrm{setting}\sqrt{{({2\mathrm{ax}}^{-1}+1)}^3}=\mathrm{u}\Rightarrow {2\mathrm{ax}}^{-1}+1={\mathrm{u}}^{\frac{2}{3}}$$$${\mathrm{x}}^{-2}\mathrm{dx}=-\frac{1}{3\mathrm{a}\sqrt[3]{\mathrm{u}}}\mathrm{du}\wedge \frac{1}{\mathrm{x}}=\frac{\sqrt[3]{{\mathrm{u}}^2}-1}{2\mathrm{a}}$$$$\mathrm{I}=\int -\left(\frac{1}{3\mathrm{a}\sqrt[3]{\mathrm{u}}}\right)\left(\frac{\sqrt[3]{{\mathrm{u}}^2}-1}{2\mathrm{a}}\right)\left(\frac{1}{\mathrm{u}}\right)\mathrm{du}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}\int \frac{({\mathrm{u}}^{\frac{2}{3}}-1)}{{\mathrm{u}}^{\frac{4}{3}}}\mathrm{du}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}\int ({\mathrm{u}}^{-\frac{2}{3}}-{\mathrm{u}}^{-\frac{4}{3}})\mathrm{du}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}(3\sqrt[3]{\mathrm{u}}+\frac{3}{\sqrt[3]{\mathrm{u}}})+\mathrm{c}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}(3\sqrt{{2\mathrm{ax}}^{-1}+1}+\frac{3}{\sqrt{{2\mathrm{ax}}^{-1}+1}})+\mathrm{c}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}\left(\frac{3\left(\frac{2\mathrm{a}+\mathrm{x}}{\mathrm{x}}\right)+3}{\sqrt{\frac{2\mathrm{a}+\mathrm{x}}{\mathrm{x}}}}\right)+\mathrm{c}$$$$\mathrm{I}=-\frac{1}{{6\mathrm{a}}^2}(\frac{6\mathrm{a}+6\mathrm{x}}{\mathrm{x}}.\frac{\sqrt{\mathrm{x}}}{\sqrt{2\mathrm{a}+\mathrm{x}}})+\mathrm{c}$$$$\mathrm{I}=-\frac{\mathrm{a}+\mathrm{x}}{{\mathrm{a}}^2}\left(\frac{1}{\sqrt{2\mathrm{ax}+{\mathrm{x}}^2}}\right)+\mathrm{c}.▴$$

Answered by ajfour last updated on 14/Nov/20

$$I=\int \frac{dx}{{(2ax+x^2)}^{3/2}}=\int \frac{dx}{{\{{(x+a)}^2-a^2\}}^{3/2}}$$$$letx+a=a\sec \theta$$$$\Rightarrow dx=a\sec \theta \tan \theta d\theta$$$$I=\int \frac{a\sec \theta \tan \theta d\theta }{a^3\tan {}^3\theta }=\frac{1}{a^2}\int \frac{\cos \theta }{\sin {}^2\theta }d\theta$$$$Nowlet\sin \theta =t\Rightarrow \cos \theta d\theta =dt$$$$\Rightarrow I=\frac{1}{a^2}\int \frac{dt}{t^2}=-\frac{1}{a^{2}t}+c=-\frac{1}{a^2\sin \theta }+c$$$$I=-\frac{1/a^2}{\sqrt{1-\frac{a^2}{{(x+a)}^2}}}+c$$$$I=\frac{-\mid x+a\mid }{a^2\sqrt{x^2+2ax}}+c.$$

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