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

The integral of (x^{1/4})/(sqrt(x)+1) is 4(1/3 x^{3/4}-\sqrt[4]{x}+arctan(\sqrt[4]{x}))+C

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{4 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

= 4 \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}

= 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

Add a constant to the solution

\int tan (s) d s

\int \frac{3}{4 x ^{2} + 9} d x

\int \frac{3}{( t ^{2} - 9 ) ^{2}} d t

\int 6 - x d x

\int x (x + 4) (2 x - 1) d x

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