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Question Number 77221 by john santu last updated on 04/Jan/20

$$\int \frac{\sqrt{2+x}dx}{\sqrt{x^5}}=?$$

Answered by jagoll last updated on 04/Jan/20

$$\int \frac{\sqrt{2+\mathrm{x}}}{{\mathrm{x}}^2\sqrt{\mathrm{x}}}\mathrm{dx}=\int \frac{1}{{\mathrm{x}}^2}\sqrt{\frac{2}{\mathrm{x}}+1}\mathrm{dx}$$$$\mathrm{let}\frac{2}{\mathrm{x}}+1={\mathrm{r}}^2\Rightarrow -\frac{2}{{\mathrm{x}}^2}\mathrm{dx}=2\mathrm{r}\mathrm{dr}$$$$\frac{1}{{\mathrm{x}}^2}\mathrm{dx}=-\mathrm{r}\mathrm{dr}$$$$\mathrm{now}\int \frac{1}{{\mathrm{x}}^2}\sqrt{\frac{2}{\mathrm{x}}+1}\mathrm{dx}=\int -{\mathrm{r}}^2\mathrm{dr}$$$$=-\frac{1}{3}{\mathrm{r}}^3+\mathrm{c}=-\frac{1}{3}\sqrt{{(\frac{2}{\mathrm{x}}+1)}^3}+\mathrm{c}$$

Commented by petrochengula last updated on 04/Jan/20

$$youandIhavebeenthinkingaboutthesamething$$

Commented by jagoll last updated on 04/Jan/20

$$\mathrm{haha}...\mathrm{yes}\mathrm{sir}$$

Commented by john santu last updated on 05/Jan/20

$$thanksboth$$

Answered by petrochengula last updated on 04/Jan/20

$$=\int \frac{\sqrt{2+x}}{x^2\sqrt{x}}dx$$$$=\int \frac{1}{x^2}\sqrt{\frac{2+x}{x}}dx$$$$=\int \frac{1}{x^2}\sqrt{\frac{2}{x}+1}dx$$$$lett=\sqrt{\frac{2}{x}+1}$$$$t^2=\frac{2}{x}+1$$$$t\frac{dt}{dx}=-\frac{1}{x^2}$$$$-tdt=\frac{1}{x^2}dx$$$$=-\int t.tdt$$$$=-\int t^{2}dt=-\frac{t^3}{3}+c=-\frac{1}{3}{(\frac{2}{x}+1)}^{\frac{3}{2}}+C$$

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