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Question Number 93720 by john santu last updated on 14/May/20

$$\int \frac{\mathit{d}\mathit{x}}{{(\mathit{x}+1)}^3\sqrt{{\mathit{x}}^2+2\mathit{x}}}$$

Commented by john santu last updated on 14/May/20

$${\mathit{x}}^2+2\mathit{x}={\mathit{t}}^2\Rightarrow (2\mathit{x}+2)\mathit{d}\mathit{x}=2\mathit{t}\mathit{d}\mathit{t}$$$$\mathit{d}\mathit{x}=\frac{\mathit{t}\mathit{d}\mathit{t}}{\mathit{x}+1};{(\mathit{x}+1)}^2={\mathit{t}}^2+1\Rightarrow \frac{1}{{(\mathit{x}+1)}^2}=\frac{1}{{\mathit{t}}^2+1}$$$$\int \frac{\mathit{d}\mathit{x}}{{(\mathit{x}+1)}^3\sqrt{{\mathit{x}}^2+2\mathit{x}}}=\int \frac{\mathit{t}\mathit{d}\mathit{t}}{{({\mathit{t}}^2+1)}^2.\mathit{t}}$$$$=\int \frac{\mathit{d}\mathit{t}}{{({\mathit{t}}^2+1)}^2}$$$$\mathrm{let}\mathrm{t}=\tan \mathrm{u}$$$$=\int \frac{\sec {}^2\mathit{u}\mathit{d}\mathit{u}}{\sec {}^4\mathit{u}}=\int {\cos }^2\mathrm{u}\mathrm{du}$$$$=\int \frac{1}{2}+\frac{1}{2}\cos 2\mathrm{u}\mathrm{du}$$$$=\frac{1}{2}\mathrm{u}+\frac{1}{2}\sin \mathrm{ucos}\mathrm{u}+\mathrm{c}$$$$=\frac{1}{2}{\tan }^{-1}\left(\mathrm{t}\right)+\frac{\mathrm{t}}{2({\mathrm{t}}^2+1)}+\mathrm{c}$$$$=\frac{1}{2}\tan {}^{-1}\left(\sqrt{{\mathrm{x}}^2+2\mathrm{x}}\right)+\frac{\sqrt{{\mathrm{x}}^2+2\mathrm{x}}}{2{(\mathrm{x}+1)}^2}+\mathrm{c}$$

Commented by mathmax by abdo last updated on 14/May/20

$$I=\int \frac{dx}{{(x+1)}^3\sqrt{x^2+2x}}wedothechangement\sqrt{x^2+2x}=t\Rightarrow$$$$x^2+2x=t^2\Rightarrow (2x+2)dx=2tdt\Rightarrow (x+1)dx=tdt\Rightarrow dx=\frac{tdt}{(x+1)}$$$$x^2+2x+1=t^2+1\Rightarrow {(x+1)}^2=(t^2+1)\Rightarrow \frac{dx}{{(x+1)}^3}=\frac{tdt}{(t^2+1)}$$$$\frac{dx}{{(x+1)}^3}=\frac{dx}{(x+1)}\times \frac{1}{{(x+1)}^2}=\frac{tdt}{{(x+1)}^2{(x+1)}^2}=\frac{tdt}{{(t^2+1)}^2}\Rightarrow$$$$I=\int \frac{tdt}{t{(t^2+1)}^2}dt=\int \frac{dt}{{(t^2+1)}^2}{=}_{t=tan\theta }\int \frac{(1+{tan}^2\theta )d\theta }{{(1+{tan}^2\theta )}^2}$$$$=\int \frac{d\theta }{1+{tan}^2\theta }=\int {cos}^2\theta d\theta =\frac{1}{2}\int (1+cos\left(2\theta \right)d\theta =\frac{1}{2}\theta +\frac{1}{4}sin\left(2\theta \right)+c$$$$=\frac{arctant}{2}+\frac{1}{4}\times \frac{2t}{1+t^2}+C=\frac{arctant}{2}+\frac{t}{2(1+t^2)}+C$$$$I=\frac{1}{2}arctan\left(\sqrt{x^2+2x}\right)+\frac{\sqrt{x^2+2x}}{2{(1+x)}^2}+C$$

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