∫1x12+x13dx=∫1x131+x16dx Put x=t6 dx=6t5dt=∫6t5dtt2(1+t)=6∫6t2t2(1+t)dt=6∫t2−t+1−11+tdt=6t33−t22+t−log⁡1+t+c=2x12−3x13+6x16−6log⁡1+x16+c

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

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$2 x^{\frac{1}{2}} - 2 x^{\frac{1}{3}} + c$

$3 x^{2} - 3 x^{3} + 6 x^{6} - 6 \log |1 + x| + c$

$2 x^{\frac{1}{2}} - 3 x^{\frac{1}{3}} + 6 x^{\frac{1}{6}} - 6 \log |1 + x^{\frac{1}{6}}| + c$

$3 x^{\frac{1}{2}} - 2 x^{\frac{1}{3}} + 6 x^{\frac{1}{6}} - 3 \log |1 + x^{\frac{1}{6}}| + c$

answer is C.

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Detailed Solution

$\int \frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}} d x = \int \frac{1}{x^{\frac{1}{3}} (1 + x^{\frac{1}{6}})} d x$

$Put x = t^{6} d x = 6 t^{5} d t$

$\begin{array}{l} = \int \frac{6 t^{5} d t}{t^{2} (1 + t)} \\ = 6 \int \frac{6 t^{2}}{t^{2} (1 + t)} d t \\ = 6 [\int (t^{2} - t + 1) - \frac{1}{1 + t}] d t \\ = 6 [\frac{t^{3}}{3} - \frac{t^{2}}{2} + t - \log |1 + t|] + c \\ = 2 x^{\frac{1}{2}} - 3 x^{\frac{1}{3}} + 6 x^{\frac{1}{6}} - 6 \log |1 + x^{\frac{1}{6}}| + c \end{array}$