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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (\sqrt[4]{x})/(sqrt(x)+1)

Solution steps

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

Simplify

Add a constant to the solution

\int 10 x e^{2 x} d x

\int \frac{cos ( x )}{sin ^{2} ( x ) + 4} d x

\int (e^{- s t} t) d t

x^{2} + 1

\int \frac{6}{3 x} d x

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