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Question Number 122798 by aurpeyz last updated on 19/Nov/20

Commented by mohammad17 last updated on 19/Nov/20

$$I={\int }_1^3\frac{1}{x^2}{(1+\frac{3}{x})}^{\frac{2}{3}}dx=-\frac{1}{5}{\left(2\right)}^{\frac{5}{3}}+\frac{1}{5}{\left(4\right)}^{\frac{5}{3}}$$

Answered by MJS_new last updated on 19/Nov/20

$$\int \frac{1}{x^2}{\left(\frac{x+3}{x}\right)}^{2/3}dx=$$$$[t={\left(\frac{x+3}{x}\right)}^{1/3}\to dx=-{(x+3)}^{2/3}{x}^{4/3}dt]$$$$=-\int t^{4}dt=-\frac{t^5}{5}=-\frac{{(x+3)}^{5/3}}{5x^{5/3}}+C$$$$\nRightarrow$$$$\underset{1}{\stackrel{3}{\int }}\frac{1}{x^2}{\left(\frac{x+3}{x}\right)}^{2/3}dx=\frac{2\sqrt[3]{2}}{5}(4-\sqrt[3]{2})$$

Answered by Dwaipayan Shikari last updated on 20/Nov/20

$${\int }_1^3\frac{1}{x^2}{\left(\frac{x+3}{x}\right)}^{\frac{2}{3}}dx$$$$=-\frac{1}{6}{\int }_1^3-\frac{6}{x^2}{(1+\frac{3}{x})}^{\frac{2}{3}}dx$$$$=\frac{1}{6}{\int }_2^4{t}^{\frac{2}{3}}dt=\frac{1}{10}{\left[t^{\frac{5}{3}}\right]}_2^4=\frac{4}{5}\sqrt[3]{2}-\frac{1}{5}\sqrt[3]{2^2}$$

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