What is the integral of x^3sqrt(81+x^2) ?

The integral of x^3sqrt(81+x^2) is 1/5 (81+x^2)^{5/2}-27(81+x^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^3sqrt(81+x^2)

\int x^{3} 81 + x^{2} d x

\frac{1}{5} (81 + x^{2})^{\frac{5}{2}} - 27 (81 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 81 + x^{2} d x

Apply u-substitution

: \int (u^{2} - 81) u^{2} d u

= \int (u^{2} - 81) u^{2} d u

Expand

(u^{2} - 81) u^{2} : u^{4} - 81 u^{2}

= \int u^{4} - 81 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 81 u^{2} d u

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

\int 81 u^{2} d u = 27 u^{3}

= \frac{u ^{5}}{5} - 27 u^{3}

Substitute back

u = 81 + x^{2}

= \frac{( 81 + x ^{2} ) ^{5}}{5} - 27 (81 + x^{2})^{3}

Simplify

\frac{( 81 + x ^{2} ) ^{5}}{5} - 27 (81 + x^{2})^{3} : \frac{1}{5} (81 + x^{2})^{\frac{5}{2}} - 27 (81 + x^{2})^{\frac{3}{2}}

= \frac{1}{5} (81 + x^{2})^{\frac{5}{2}} - 27 (81 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (81 + x^{2})^{\frac{5}{2}} - 27 (81 + x^{2})^{\frac{3}{2}} + C

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

x \to 0 lim (\frac{4}{14 + x})

y = \frac{( x - 2 )}{y + 3}

f (x) = e^{x^{2} + 1}

x \to 0 lim (1 + \frac{1}{4 x ^{2}})

What is the integral of x^3sqrt(81+x^2) ?
The integral of x^3sqrt(81+x^2) is 1/5 (81+x^2)^{5/2}-27(81+x^2)^{3/2}+C