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

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(A) To solve the integral $I = \int \frac{dx}{x^{1/2} + x^{1/3}}$,we use the substitution $x = t^6$,so $dx = 6t^5 dt$.Substituting these into the inte

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$2, -3, 6, -6$

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(A) To solve the integral $I = \int \frac{dx}{x^{1/2} + x^{1/3}}$,we use the substitution $x = t^6$,so $dx = 6t^5 dt$.Substituting these into the integral,we get:$I = \int \frac{6t^5 dt}{t^3 + t^2} = \int \frac{6t^5}{t^2(t+1)} dt = \int \frac{6t^3}{t+1} dt$.Using polynomial division,$\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$.Thus,$I = 6 \int (t^2 - t + 1 - \frac{1}{t+1}) dt = 6 [\frac{t^3}{3} - \frac{t^2}{2} + t - \log|t+1|] + k$.$I = 2t^3 - 3t^2 + 6t - 6 \log|t+1| + k$.Substituting $t = x^{1/6}$ back,we get:$I = 2(x^{1/6})^3 - 3(x^{1/6})^2 + 6(x^{1/6}) - 6 \log(x^{1/6} + 1) + k$.$I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log(x^{1/6} + 1) + k$.Comparing this with the given form $A x^{1/2} + B x^{1/3} + C x^{1/6} + D \log(x^{1/6} + 1) + k$,we find $A = 2, B = -3, C = 6, D = -6$.

$\int \frac{d x}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=A x^{\frac{1}{2}}+B x^{\frac{1}{3}}+C x^{\frac{1}{6}}+D \log \left(x^{\frac{1}{6}}+1\right)+k$ (where $k$ is the integration constant),then the values of $A, B, C$ and $D$ are respectively,

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