What is the integral of x^5sqrt(x^3+4) ?

The integral of x^5sqrt(x^3+4) is 1/3 (2/5 (x^3+4)^{5/2}-8/3 (x^3+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^5sqrt(x^3+4)

\int x^{5} x^{3} + 4 d x

\frac{1}{3} (\frac{2}{5} (x^{3} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{3} + 4)^{\frac{3}{2}}) + C

Solution steps

\int x^{5} x^{3} + 4 d x

Apply u-substitution

: \int \frac{( u - 4 ) u}{3} d u

= \int \frac{( u - 4 ) 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 - 4) u d u

Expand

(u - 4) u : u^{\frac{3}{2}} - 4 u

= \frac{1}{3} \cdot \int u^{\frac{3}{2}} - 4 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 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = x^{3} + 4

= \frac{1}{3} (\frac{2}{5} (x^{3} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{3} + 4)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (x^{3} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{3} + 4)^{\frac{3}{2}}) + C

\int \frac{( x + 5 )}{x ^{2} + 10 x + 26} d x

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

\int sin^{4} (2 x) \cdot cos (2 x) d x

y = \frac{1}{( 1 + c _{1} e ^{- x} )}

\frac{\partial}{\partial x} (e^{x t})

What is the integral of x^5sqrt(x^3+4) ?
The integral of x^5sqrt(x^3+4) is 1/3 (2/5 (x^3+4)^{5/2}-8/3 (x^3+4)^{3/2})+C