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Question Number 22545 by vajpaithegrate@gmail.com last updated on 20/Oct/17

$$\int \frac{{\mathrm{x}}^{\frac{1}{2}}}{{\mathrm{x}}^{\frac{1}{2}}-{\mathrm{x}}^{\frac{1}{3}}}\mathrm{dx}=$$$$$$

Answered by $@ty@m last updated on 20/Oct/17

$$ItssimilartoQ.No.20540$$

Commented by vajpaithegrate@gmail.com last updated on 20/Oct/17

$$\mathrm{Thank}\mathrm{u}\mathrm{sir}$$

Answered by sma3l2996 last updated on 20/Oct/17

$$I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}}dx=\int \frac{x}{x-\sqrt[3]{x^2}}dx$$$$u=\sqrt[3]{x}\iff x=u^3\Rightarrow dx=3u^{2}du$$$$I=\int \frac{u^3}{u^3-u^2}\left(3u^{2}du\right)$$$$I=3\int \frac{u^3}{u-1}du=3\int \frac{u^3-1+1}{u-1}du$$$$=3\int \frac{(u-1)(u^2+u+1)}{u-1}du+3\int \frac{du}{u-1}$$$$=3\int (u^2+u+1)du+3ln\mid u-1\mid +c$$$$I=u^3+\frac{1}{2}{u}^2+3u+3ln\mid u-1\mid +C$$$$I=x+\frac{3}{2}\sqrt[3]{x^2}+3\sqrt[3]{x}+3ln\mid \sqrt[3]{x}-1\mid +C$$

Commented by vajpaithegrate@gmail.com last updated on 21/Oct/17

$$\int \frac{\sqrt{\mathrm{x}}}{\sqrt{}\mathrm{x}{-}^3\sqrt{\mathrm{x}}}\mathrm{dx}=\int \frac{\mathrm{x}}{\mathrm{x}{-}^3\sqrt{{\mathrm{x}}^2}}\mathrm{dx}\mathrm{is}\mathrm{it}\mathrm{correc}$$

Commented by $@ty@m last updated on 21/Oct/17

$${}^3\sqrt{x}\times \sqrt{x}$$$$=x^{\frac{1}{3}}\times x^{\frac{1}{2}}$$$$=x^{\frac{5}{6}}$$$${}^3\sqrt{x^2}=x^{\frac{2}{3}}$$$$\therefore {}^3\sqrt{x}\times \sqrt{x}\ne {}^3\sqrt{x^2}$$

Commented by vajpaithegrate@gmail.com last updated on 21/Oct/17

$$\mathrm{yes}\mathrm{sir}$$

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