The value of ∫dxx+x3 is

# The value of $\int \frac{dx}{\sqrt{x}+\sqrt[3]{x}}$ is

1. A

$3\sqrt{x}+3\left(\sqrt[3]{x}\right)-6\left(\sqrt[6]{x}\right)+6\mathrm{log}\left(\sqrt[6]{x}+1\right)+C$

2. B

$2\sqrt{x}+6\sqrt[6]{x}-6\mathrm{log}\left(\sqrt[6]{x}+1\right)+C$

3. C

$2\sqrt{x}-3\left(\sqrt[3]{x}\right)+6\left(\sqrt[6]{x}\right)-6\mathrm{log}\left(\sqrt[6]{x}+1\right)+C$

4. D

none of these

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### Solution:

We want a substitution that will allow us to

find both the square root and the cube root without getting

fractional exponents. Thus we want a substitution of the

form $x={u}^{k}$ where $k$ is a multiple of 2 and 3. Let us use the
least common multiple 6.

Let $x={u}^{6},dx=6{u}^{5}du$

$\begin{array}{l}\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt[3]{x}}=\int \frac{6{u}^{5}du}{{u}^{3}+{u}^{2}}=6\int \frac{{u}^{3}}{u+1}du\\ =6\int \left({u}^{2}-u+1-\frac{1}{u+1}\right)du\\ =2{u}^{3}-3{u}^{2}+6u-6\mathrm{log}\left(u+1\right)+C\\ =2\sqrt{x}-3\left(\sqrt[3]{x}\right)+6\left(\sqrt[6]{x}\right)-6\mathrm{log}\left(\sqrt[6]{x}+1\right)+C\end{array}$

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