What is the integral of x^5sqrt(x^3+2) ?

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

\int x^{5} x^{3} + 2 d x

\frac{1}{3} (\frac{2}{5} (x^{3} + 2)^{\frac{5}{2}} - \frac{4}{3} (x^{3} + 2)^{\frac{3}{2}}) + C

Solution steps

\int x^{5} x^{3} + 2 d x

Apply u-substitution

: \int \frac{( u - 2 ) u}{3} d u

= \int \frac{( u - 2 ) u}{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 - 2) u d u

Expand

(u - 2) u : u^{\frac{3}{2}} - 2 u

= \frac{1}{3} \cdot \int u^{\frac{3}{2}} - 2 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{1}{3} (\int u^{\frac{3}{2}} d u - \int 2 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = x^{3} + 2

= \frac{1}{3} (\frac{2}{5} (x^{3} + 2)^{\frac{5}{2}} - \frac{4}{3} (x^{3} + 2)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (x^{3} + 2)^{\frac{5}{2}} - \frac{4}{3} (x^{3} + 2)^{\frac{3}{2}}) + C

\frac{d}{d x} (\frac{2 - 5 x - x ^{2}}{x ^{2} - 6})

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

\frac{d y}{d x} = 2 x \cdot e^{- y}

\int \frac{x}{x - 10} d x

n = 1 \sum \infty n 2

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