What is the integral of x^3sqrt(12+x^4) ?

The integral of x^3sqrt(12+x^4) is 1/6 (12+x^4)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^3sqrt(12+x^4)

\int x^{3} 12 + x^{4} d x

\frac{1}{6} (12 + x^{4})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 12 + x^{4} d x

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 12 + x^{4}

= \frac{1}{4} \cdot \frac{2}{3} (12 + x^{4})^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (12 + x^{4})^{\frac{3}{2}} : \frac{1}{6} (12 + x^{4})^{\frac{3}{2}}

= \frac{1}{6} (12 + x^{4})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (12 + x^{4})^{\frac{3}{2}} + C

n = 1 \sum \infty (- 1)^{n}

\frac{d}{d x} (9 x - 5)

\int (x - 8)^{\frac{3}{2}} d x

\frac{d}{d x} (e^{2 x y})

f (x) = x^{2} (1 - 4 x)

What is the integral of x^3sqrt(12+x^4) ?
The integral of x^3sqrt(12+x^4) is 1/6 (12+x^4)^{3/2}+C