What is the integral of (x^{1/2})/(x+x^{2/3)} ?

The integral of (x^{1/2})/(x+x^{2/3)} is 6(1/3 x^{1/2}-x^{1/6}+arctan(x^{1/6}))+C

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integral of (x^{1/2})/(x+x^{2/3)}

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

6 (\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}})) + C

Solution steps

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

Apply u-substitution

: \int \frac{6 u ^{4}}{u ^{2} + 1} d u

= \int \frac{6 u ^{4}}{u ^{2} + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{u ^{4}}{u ^{2} + 1} d u

Long division

\frac{u ^{4}}{u ^{2} + 1} : u^{2} - 1 + \frac{1}{u ^{2} + 1}

= 6 \cdot \int u^{2} - 1 + \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int u^{2} d u - \int 1 d u + \int \frac{1}{u ^{2} + 1} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 6 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

u = x^{\frac{1}{6}}

= 6 \frac{( x ^{\frac{1}{6}} ) ^{3}}{3} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}})

Simplify

6 \frac{( x ^{\frac{1}{6}} ) ^{3}}{3} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}}) : 6 (\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}}))

= 6 (\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}}))

Add a constant to the solution

= 6 (\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{1}{6}} + arctan (x^{\frac{1}{6}})) + C

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What is the integral of (x^{1/2})/(x+x^{2/3)} ?
The integral of (x^{1/2})/(x+x^{2/3)} is 6(1/3 x^{1/2}-x^{1/6}+arctan(x^{1/6}))+C