int frac{1}{x^{frac{1}{2}}+x^{frac{1}{3}}} , | vedclass

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(B) Let $I = \int \frac{dx}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}$.Substitute $x^{\frac{1}{6}} = t$,then $x = t^6$ and $dx = 6t^5 \, dt$.Also,$x^{\frac{1

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$2 \sqrt{x}-3 \sqrt[3]{x}+6 \sqrt[6]{x}-6 \log |\sqrt[6]{x}+1|+c$

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(B) Let $I = \int \frac{dx}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}$.Substitute $x^{\frac{1}{6}} = t$,then $x = t^6$ and $dx = 6t^5 \, dt$.Also,$x^{\frac{1}{2}} = t^3$ and $x^{\frac{1}{3}} = t^2$.Substituting these into the integral:$I = \int \frac{6t^5 \, dt}{t^3 + t^2} = 6 \int \frac{t^5}{t^2(t+1)} \, dt = 6 \int \frac{t^3}{t+1} \, dt$.Using polynomial division or algebraic manipulation:$I = 6 \int \frac{(t^3 + 1) - 1}{t+1} \, dt = 6 \int \left( \frac{(t+1)(t^2 - t + 1)}{t+1} - \frac{1}{t+1} \right) \, dt$.$I = 6 \int (t^2 - t + 1) \, dt - 6 \int \frac{1}{t+1} \, dt$.$I = 6 \left( \frac{t^3}{3} - \frac{t^2}{2} + t \right) - 6 \log |t+1| + c$.$I = 2t^3 - 3t^2 + 6t - 6 \log |t+1| + c$.Substituting back $t = x^{\frac{1}{6}}$:$I = 2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \log |\sqrt[6]{x} + 1| + c$.

$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} \, dx =$

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