What is the integral of x^3sqrt(x^4+9) ?

The integral of x^3sqrt(x^4+9) is 1/6 (x^4+9)^{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(x^4+9)

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

\frac{1}{6} (x^{4} + 9)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{4} + 9 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 = x^{4} + 9

= \frac{1}{4} \cdot \frac{2}{3} (x^{4} + 9)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (x^{4} + 9)^{\frac{3}{2}} : \frac{1}{6} (x^{4} + 9)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} + 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (x^{4} + 9)^{\frac{3}{2}} + C

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

\frac{\partial}{\partial y} (2 x y^{2} - 3)

x \to 1 lim (\frac{x + 4}{x + 2})

\frac{d}{d x} (4 + 3 x)

\int cos (\frac{mπ x}{2}) d x

What is the integral of x^3sqrt(x^4+9) ?
The integral of x^3sqrt(x^4+9) is 1/6 (x^4+9)^{3/2}+C