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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x))/(\sqrt[4]{x^3)+1}

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

Solution steps

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{4 ( u - 1 )}{3 u} d u

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

Take the constant out:

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{4}{3} \cdot \int 1 - \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{4}{3} (u - ln ∣ u ∣)

Substitute back

u = x^{\frac{3}{4}} + 1

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

Add a constant to the solution

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

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

\int 2 sin^{5} (x) cos (x) d x

\int cos^{2} (ω t + φ) sin (ω t + φ) d t

\int 1 d sin (x) d x

\int x (4 - x)^{5} d x

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