Let I=∫dxx+x3 Put x=t6 differentiate w.r.to 't' on both sides I=∫6t5t3+t2dt =6∫t3t+1dt =6∫t3+1-1t+1dt =6∫(t2-t+1-1t+1) dt =6t33-t22+t-log(t+1)

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$\int \frac{d x}{\sqrt{x} + \sqrt[3]{x}} =$

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

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

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

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

answer is D.

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Detailed Solution

$L e t I = \int \frac{d x}{\sqrt{x} + \sqrt[3]{x}} Put x = t^{6} d i f f e r e n t i a t e w . r . t o' t' o n b o t h s i d e s I = \int \frac{6 t^{5}}{t^{3} + t^{2}} d t = 6 \int \frac{t^{3}}{t + 1} d t = 6 \int \frac{t^{3} + 1 - 1}{t + 1} d t = 6 \int ([t^{2} - t + 1] - \frac{1}{t + 1}) d t = 6 [\frac{t^{3}}{3} - \frac{t^{2}}{2} + t] - \log (t + 1)$