What is the integral of (x^{1/2})/(x+x^{1/4)} ?

The integral of (x^{1/2})/(x+x^{1/4)} is

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integral of (x^{1/2})/(x+x^{1/4)}

\int \frac{x ^{\frac{1}{2}}}{x + x ^{\frac{1}{4}}} d x

4 (\frac{1}{2} x^{\frac{1}{2}} + \frac{1}{3} ln x^{\frac{1}{4}} + 1 - \frac{1}{6} ln 4 x^{\frac{1}{2}} - 4 x^{\frac{1}{4}} + 4 - \frac{1}{3 ^{\frac{1}{2}}} arctan (\frac{1}{3} (2 x^{\frac{1}{4}} - 1))) + C

Solution steps

\int \frac{x ^{\frac{1}{2}}}{x + x ^{\frac{1}{4}}} d x

Apply u-substitution

: \int \frac{4 u ^{4}}{u ^{3} + 1} d u

= \int \frac{4 u ^{4}}{u ^{3} + 1} d u

Take the constant out:

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

= 4 \cdot \int \frac{u ^{4}}{u ^{3} + 1} d u

Long division

\frac{u ^{4}}{u ^{3} + 1} : u - \frac{u}{u ^{3} + 1}

= 4 \cdot \int u - \frac{u}{u ^{3} + 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

= 4 (\int u d u - \int \frac{u}{u ^{3} + 1} d u)

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

\int \frac{u}{u ^{3} + 1} d u = - \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

= 4 (\frac{u ^{2}}{2} - (- \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))))

Substitute back

u = x^{\frac{1}{4}}

= 4 \frac{( x ^{\frac{1}{4}} ) ^{2}}{2} - (- \frac{1}{3} ln x^{\frac{1}{4}} + 1 + \frac{1}{6} (ln 4 (x^{\frac{1}{4}})^{2} - 4 x^{\frac{1}{4}} + 4 + 23 arctan (\frac{2 x ^{\frac{1}{4}} - 1}{3})))

Simplify

4 \frac{( x ^{\frac{1}{4}} ) ^{2}}{2} - (- \frac{1}{3} ln x^{\frac{1}{4}} + 1 + \frac{1}{6} (ln 4 (x^{\frac{1}{4}})^{2} - 4 x^{\frac{1}{4}} + 4 + 23 arctan (\frac{2 x ^{\frac{1}{4}} - 1}{3}))) : 4 (\frac{1}{2} x^{\frac{1}{2}} + \frac{1}{3} ln x^{\frac{1}{4}} + 1 - \frac{1}{6} ln 4 x^{\frac{1}{2}} - 4 x^{\frac{1}{4}} + 4 - \frac{1}{3 ^{\frac{1}{2}}} arctan (\frac{1}{3} (2 x^{\frac{1}{4}} - 1)))

= 4 (\frac{1}{2} x^{\frac{1}{2}} + \frac{1}{3} ln x^{\frac{1}{4}} + 1 - \frac{1}{6} ln 4 x^{\frac{1}{2}} - 4 x^{\frac{1}{4}} + 4 - \frac{1}{3 ^{\frac{1}{2}}} arctan (\frac{1}{3} (2 x^{\frac{1}{4}} - 1)))

Add a constant to the solution

= 4 (\frac{1}{2} x^{\frac{1}{2}} + \frac{1}{3} ln x^{\frac{1}{4}} + 1 - \frac{1}{6} ln 4 x^{\frac{1}{2}} - 4 x^{\frac{1}{4}} + 4 - \frac{1}{3 ^{\frac{1}{2}}} arctan (\frac{1}{3} (2 x^{\frac{1}{4}} - 1))) + C

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What is the integral of (x^{1/2})/(x+x^{1/4)} ?
The integral of (x^{1/2})/(x+x^{1/4)} is