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

The integral of x^2(x^3+1)^{1/4} is 4/15 (x^3+1)^{5/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2(x^3+1)^{1/4}

\int x^{2} (x^{3} + 1)^{\frac{1}{4}} d x

\frac{4}{15} (x^{3} + 1)^{\frac{5}{4}} + C

Solution steps

\int x^{2} (x^{3} + 1)^{\frac{1}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= \frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = x^{3} + 1

= \frac{1}{3} \cdot \frac{4}{5} (x^{3} + 1)^{\frac{5}{4}}

Simplify

\frac{1}{3} \cdot \frac{4}{5} (x^{3} + 1)^{\frac{5}{4}} : \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}}

= \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{15} (x^{3} + 1)^{\frac{5}{4}} + C

\frac{\partial}{\partial t} (2 cos (t))

f (x) = 2 arctan (x)

\frac{d}{d y} (\frac{1}{y})

\int - cos^{3} (x) d x

\frac{d}{d x} (sin (3) x^{2} cos (2 x))

What is the integral of x^2(x^3+1)^{1/4} ?
The integral of x^2(x^3+1)^{1/4} is 4/15 (x^3+1)^{5/4}+C