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Question Number 116701 by bemath last updated on 06/Oct/20

$$\int \frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}}=?$$

Commented by bobhans last updated on 06/Oct/20

$$\int \frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}}=\int \frac{\mathrm{dx}}{\mathrm{x}(1+\sqrt{\mathrm{x}})}$$$$[\mathrm{letting}\mathrm{x}={\lambda }^2\to \mathrm{dx}=2\lambda \mathrm{d}\lambda ]$$$$\mathrm{I}=\int \frac{2\lambda \mathrm{d}\lambda }{{\lambda }^2(1+\lambda )}=\int \frac{2\mathrm{d}\lambda }{\lambda (1+\lambda )}$$$$=2[\int (\frac{1}{\lambda }-\frac{1}{1+\lambda })\mathrm{d}\lambda ]$$$$=2[\ln \lambda -\ln (1+\lambda )]+\mathrm{c}$$$$=2\ln \left(\frac{\sqrt{\mathrm{x}}}{1+\sqrt{\mathrm{x}}}\right)+\mathrm{c}$$

Commented by MJS_new last updated on 06/Oct/20

$$\int \frac{dx}{x(1+x^q)}=-\frac{1}{q}\ln \mid \frac{1}{x^q}+1\mid +C=\ln \mid x\mid -\frac{1}{q}\ln \mid x^q+1\mid +C$$

Answered by Dwaipayan Shikari last updated on 06/Oct/20

$$\int \frac{dx}{x+x\sqrt{x}}x=t^2\Rightarrow 1=2t\frac{dt}{dx}$$$$\int \frac{2tdt}{t^2(1+t)}=2\int \frac{1}{t(1+t)}dt=2log\left(\frac{t}{t+1}\right)+C$$$$=2log\left(\frac{\sqrt{x}}{\sqrt{x}+1}\right)+C$$

Commented by bemath last updated on 06/Oct/20

$$\mathrm{gaves}\mathrm{kudos}$$$$$$

Answered by mathmax by abdo last updated on 06/Oct/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{x}}}\Rightarrow \mathrm{I}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}\int \frac{2\mathrm{tdt}}{{\mathrm{t}}^2+{\mathrm{t}}^3}=2\int \frac{\mathrm{dt}}{\mathrm{t}+{\mathrm{t}}^2}$$$$=2\int \frac{\mathrm{dt}}{\mathrm{t}(\mathrm{t}+1)}=2\int (\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}+1})\mathrm{dt}=2\ln \mid \frac{\mathrm{t}}{\mathrm{t}+1}\mid +\mathrm{c}$$$$=2\ln \mid \frac{\sqrt{\mathrm{x}}}{1+\sqrt{\mathrm{x}}}\mid +\mathrm{c}$$

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