What is the integral of x^5sqrt(x^6+8) ?

The integral of x^5sqrt(x^6+8) is 1/9 (x^6+8)^{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^5sqrt(x^6+8)

\int x^{5} x^{6} + 8 d x

\frac{1}{9} (x^{6} + 8)^{\frac{3}{2}} + C

Solution steps

\int x^{5} x^{6} + 8 d x

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{6} + 8

= \frac{1}{6} \cdot \frac{2}{3} (x^{6} + 8)^{\frac{3}{2}}

Simplify

\frac{1}{6} \cdot \frac{2}{3} (x^{6} + 8)^{\frac{3}{2}} : \frac{1}{9} (x^{6} + 8)^{\frac{3}{2}}

= \frac{1}{9} (x^{6} + 8)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} (x^{6} + 8)^{\frac{3}{2}} + C

\int \frac{ln ( x )}{x ^{18}} d x

\int (x^{2} + 2) d x

\frac{\partial}{\partial y} (- cos (x y))

x \to 0 lim ((e^{x} - 1) ln (x))

\frac{d y}{d x} = (x y)^{\frac{1}{3}}

What is the integral of x^5sqrt(x^6+8) ?
The integral of x^5sqrt(x^6+8) is 1/9 (x^6+8)^{3/2}+C