$\int \frac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} d x$ is equal to

Moderate

A

$x - \frac{6}{5} x^{5 / 6} - \frac{3}{2} x^{2 / 3} - 2 x^{1 / 2} + 3 x^{1 / 3} + 6 x^{1 / 6} + 6 \log (1 + x^{1 / 6}) + C$
B

$x - \frac{6}{5} x^{5 / 6} + \frac{3}{2} x^{2 / 3} - 2 x^{1 / 2} + 3 x^{1 / 3} - 6 x^{1 / 6} + 6 \log (1 + x^{1 / 6}) + C$
C

$x + \frac{6}{5} x^{5 / 6} + \frac{3}{2} x^{2 / 3} - 2 x^{1 / 2} + 3 x^{1 / 3} - 6 x^{1 / 6} + 6 \log (1 + x^{1 / 6}) + C$
D

None of these

Solution

Put $x = z^{6} \Rightarrow d x = 6 z^{5} d z .$
$∴ \int \frac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} d x = \int \frac{z^{3} 6 z^{5} d z}{z^{3} + z^{2}} = 6 \int \frac{z^{6} d z}{z + 1}$

$= 6 \int (\frac{z^{6} - 1}{z + 1} + \frac{1}{z + 1}) d z$

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

$= 6 \int (z^{5} - z^{4} + z^{3} - z^{2} + z - 1) d z + 6 \log (z + 1)$

$= z^{6} - \frac{6}{z} z^{5} + \frac{3}{2} z^{4} - 2 z^{3} + 3 z^{2} - 6 z + 6 \log (z + 1) + C$

$= x - \frac{6}{5} x^{5 / 6} + \frac{3}{2} x^{2 / 3} - 2 x^{1 / 2} + 3 x^{1 / 3} - 6 x^{1 / 6} + 6 \log (1 + x^{1 / 6}) + C .$

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