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Question Number 129855 by liberty last updated on 20/Jan/21

$$\vartheta =\int \frac{dx}{{(1+\sqrt{x})}^3}$$

Answered by EDWIN88 last updated on 20/Jan/21

$$\vartheta =\int \frac{\mathrm{dx}}{{\left(\sqrt{\mathrm{x}}\right)}^3{(1+{\mathrm{x}}^{-1/2})}^3}=\int \frac{{\mathrm{x}}^{-3/2}}{{(1+{\mathrm{x}}^{-1/2})}^3}\mathrm{dx}$$$$\mathrm{let}\phi =1+{\mathrm{x}}^{-1/2}\to \mathrm{d}\phi =-\frac{1}{2}{\mathrm{x}}^{-3/2}\mathrm{dx}$$$$\vartheta =-2\int \frac{\mathrm{d}\phi }{{\phi }^3}=-2\int {\phi }^{-3}\mathrm{d}\phi$$$$\phi =\frac{1}{{\phi }^2}+\mathrm{C}=\frac{1}{{\left(\frac{\sqrt{\mathrm{x}}+1}{\sqrt{\mathrm{x}}}\right)}^2}+\mathrm{C}$$$$\phi =\frac{\mathrm{x}}{{(\sqrt{\mathrm{x}}+1)}^2}+\mathrm{C}$$

Answered by stelor last updated on 20/Jan/21

$$\mathrm{let}u=\sqrt{x}so,du=\frac{1}{2\sqrt{x}}dx=du=\frac{1}{2u}dx$$$$v=\int \frac{2\mathrm{udu}}{{(1+\mathrm{u})}^3}whereu=\sqrt{x}$$$$lett=1+uso,\mathrm{du}=\mathrm{dt}$$$$v=2\int \frac{(t-1)}{t^3}dtwheret=u-1=\sqrt{x}-1$$$$v=2[\int (\frac{1}{{\mathrm{t}}^2}-\frac{1}{{\mathrm{t}}^3})dt]=2(-\frac{1}{t}+\frac{1}{2t^2}+c)$$$$v=2(-\frac{1}{\sqrt{x}-1}+\frac{1}{2{(\sqrt{x}-1)}^2}+c)$$$$v=\frac{3-2\sqrt{x}}{{(\sqrt{x}-1)}^2}+c$$

Commented by bemath last updated on 20/Jan/21

$$\mathrm{t}=1+\mathrm{u}\mathrm{but}\mathrm{t}=\mathrm{u}-1;\mathrm{ambiguous}$$

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