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Question Number 131064 by shaker last updated on 01/Feb/21

Answered by Dwaipayan Shikari last updated on 01/Feb/21

$$\frac{1}{4}\int \frac{1}{x}(\frac{1}{x^2}-\frac{1}{x^2+4})dx$$$$=-\frac{1}{8x^2}-\frac{1}{16}\int \frac{1}{x}-\frac{x}{x^2+4}dx$$$$=-\frac{1}{8x^2}+\frac{1}{16}log\left(\frac{\sqrt{x^2+4}}{x}\right)+C$$

Answered by benjo_mathlover last updated on 01/Feb/21

$$\int \frac{{\mathrm{x}}^{-3}}{{\mathrm{x}}^2(1+{4\mathrm{x}}^{-2})}\mathrm{dx}$$$$\mathrm{let}\mathrm{u}=1+{4\mathrm{x}}^{-2}\Rightarrow \mathrm{du}=-{8\mathrm{x}}^{-3}\mathrm{dx}\mathrm{or}{\mathrm{x}}^{-3}\mathrm{dx}=-\frac{\mathrm{du}}{8}$$$$\mathrm{and}\frac{4}{{\mathrm{x}}^2}=\mathrm{u}-1\Rightarrow \frac{1}{{\mathrm{x}}^2}=\frac{\mathrm{u}-1}{4}$$$$\mathrm{then}\mathrm{we}\mathrm{has}\mathrm{I}=\int \frac{1}{\mathrm{u}}.\frac{\mathrm{u}-1}{4}.(-\frac{\mathrm{du}}{8})$$$$\mathrm{I}=-\frac{1}{32}\int \frac{\mathrm{u}-1}{\mathrm{u}}\mathrm{du}=-\frac{1}{32}(\mathrm{u}-\ln \mid \mathrm{u}\mid )+\mathrm{c}$$$$\mathrm{I}=-\frac{1}{32}(\frac{{\mathrm{x}}^2+4}{{\mathrm{x}}^2}-\ln ({\mathrm{x}}^2+4)+2\ln \mid \mathrm{x}\mid )+\mathrm{c}$$

Answered by Ar Brandon last updated on 01/Feb/21

$$\mathcal{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^3({\mathrm{x}}^2+4)},\mathrm{x}=2\tan \theta$$$$=\int \frac{{2\sec }^2\theta }{{8\tan }^3\theta ({4\tan }^2\theta +4)}\mathrm{d}\theta$$$$=\frac{1}{16}\int \frac{\mathrm{d}\theta }{{\tan }^3\theta }=\frac{1}{16}\int \frac{(1-{\sin }^2\theta )}{{\sin }^3\theta }\mathrm{d}\left(\sin \theta \right)$$$$=-\frac{1}{{32\sin }^2\theta }-\frac{\ln \mid \sin \theta \mid }{16}+\mathcal{C},\theta ={\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)$$

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