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

The integral of (x^{1/4})/(1+x^{1/2)} is 4(1/3 x^{3/4}-x^{1/4}+arctan(x^{1/4}))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Long division

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

= 4 \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

= 4 (\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)

= 4 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

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

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

Simplify

4 \frac{( x ^{\frac{1}{4}} ) ^{3}}{3} - x^{\frac{1}{4}} + arctan (x^{\frac{1}{4}}) : 4 (\frac{1}{3} x^{\frac{3}{4}} - x^{\frac{1}{4}} + arctan (x^{\frac{1}{4}}))

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

Add a constant to the solution

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

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\frac{d y}{d x} + 3 y = \frac{3}{2}

\frac{d}{d x} (e^{ik x})

What is the integral of (x^{1/4})/(1+x^{1/2)} ?
The integral of (x^{1/4})/(1+x^{1/2)} is 4(1/3 x^{3/4}-x^{1/4}+arctan(x^{1/4}))+C