If int frac{sqrt{x}}{sqrt{x}-sqrt[3]{x}} d x= | vedclass

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(A) Let $I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x$. Substitute $x=t^6$, so $d x=6 t^5 d t$. Then $I=\int \frac{t^3}{t^3-t^2} \cdot 6 t^5 d t =

Vedclass · Accepted Answer

$\frac{137}{10}$

Vedclass · Answer

(A) Let $I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x$. Substitute $x=t^6$, so $d x=6 t^5 d t$. Then $I=\int \frac{t^3}{t^3-t^2} \cdot 6 t^5 d t = 6 \int \frac{t^8}{t^2(t-1)} d t = 6 \int \frac{t^6}{t-1} d t$. Using polynomial division, $\frac{t^6}{t-1} = t^5+t^4+t^3+t^2+t+1+\frac{1}{t-1}$. Integrating term by term: $I = 6 \left[ \frac{t^6}{6} + \frac{t^5}{5} + \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} + t + \log|t-1| \right] + K$. $I = t^6 + \frac{6}{5} t^5 + \frac{6}{4} t^4 + \frac{6}{3} t^3 + \frac{6}{2} t^2 + 6t + 6 \log|t-1| + K$. Since $t = x^{1/6}$, $I = x + \frac{6}{5} x^{5/6} + \frac{3}{2} x^{2/3} + 2 x^{1/2} + 3 x^{1/3} + 6 x^{1/6} + \log(\sqrt[6]{x}-1)^6 + K$. Comparing coefficients: $E = \frac{6}{5}$, $D = \frac{3}{2}$, $C = 2$, $B = 3$, $A = 6$. Sum $A+B+C+D+E = 6 + 3 + 2 + \frac{3}{2} + \frac{6}{5} = 11 + 1.5 + 1.2 = 13.7 = \frac{137}{10}$.

If $\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x=x+E x^{5 / 6}+D x^{2 / 3}+C x^{1 / 2}+B x^{1 / 3}+A x^{1 / 6}+\log (\sqrt[6]{x}-1)^6+K$, then $A+B+C+D+E=$

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