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Question Number 132647 by aurpeyz last updated on 15/Feb/21

Answered by Dwaipayan Shikari last updated on 15/Feb/21

$$\int \frac{1}{\sqrt{x}(\sqrt{x}+7)}dxx=t^2$$$$=2\int \frac{1}{t+7}dt=2log(t+7)+C=log{(\sqrt{x}+7)}^2+C$$

Answered by EDWIN88 last updated on 15/Feb/21

$$\mathrm{let}\sqrt{\mathrm{x}}+7=\mathrm{u}\mathrm{and}\mathrm{x}={(\mathrm{u}-7)}^2$$$$\mathrm{I}=\int \frac{2(\mathrm{u}-7)}{(\mathrm{u}-7)\mathrm{u}}\mathrm{du}=\int \frac{2}{\mathrm{u}}\mathrm{du}=2\ln \mathrm{u}+\mathrm{c}$$$$\mathrm{I}=2\ln (\sqrt{\mathrm{x}}+7)+\mathrm{c}$$

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