What is the integral of x^3sqrt(x^2+18) ?

The integral of x^3sqrt(x^2+18) is 1/5 (x^2+18)^{5/2}-6(x^2+18)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^3sqrt(x^2+18)

\int x^{3} x^{2} + 18 d x

\frac{1}{5} (x^{2} + 18)^{\frac{5}{2}} - 6 (x^{2} + 18)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 18 d x

Apply u-substitution

= \int (u^{2} - 18) u^{2} d u

Expand

(u^{2} - 18) u^{2} : u^{4} - 18 u^{2}

= \int u^{4} - 18 u^{2} 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

= \int u^{4} d u - \int 18 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 18 u^{2} d u = 6 u^{3}

= \frac{u ^{5}}{5} - 6 u^{3}

Substitute back

u = x^{2} + 18

= \frac{( x ^{2} + 18 ) ^{5}}{5} - 6 (x^{2} + 18)^{3}

Simplify

\frac{( x ^{2} + 18 ) ^{5}}{5} - 6 (x^{2} + 18)^{3} : \frac{1}{5} (x^{2} + 18)^{\frac{5}{2}} - 6 (x^{2} + 18)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 18)^{\frac{5}{2}} - 6 (x^{2} + 18)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 18)^{\frac{5}{2}} - 6 (x^{2} + 18)^{\frac{3}{2}} + C

\frac{d}{d x} (2 cos (cot (2 x + 1)))

x \to - 7 + lim (\frac{1}{x + 7})

\int_{2}^{4} 3 e^{3 x} d x

\frac{d}{d x} (6 e^{x})

x \to 1 lim (3 x + 1)

What is the integral of x^3sqrt(x^2+18) ?
The integral of x^3sqrt(x^2+18) is 1/5 (x^2+18)^{5/2}-6(x^2+18)^{3/2}+C