What is the integral of x^3sqrt(16+x^2) ?

The integral of x^3sqrt(16+x^2) is 1/5 (16+x^2)^{5/2}-16/3 (16+x^2)^{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(16+x^2)

\int x^{3} 16 + x^{2} d x

\frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 16 + x^{2} d x

Apply u-substitution

= \int (u^{2} - 16) u^{2} d u

Expand

(u^{2} - 16) u^{2} : u^{4} - 16 u^{2}

= \int u^{4} - 16 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 16 u^{2} d u

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

\int 16 u^{2} d u = \frac{16 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{16 u ^{3}}{3}

Substitute back

u = 16 + x^{2}

= \frac{( 16 + x ^{2} ) ^{5}}{5} - \frac{16 ( 16 + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 16 + x ^{2} ) ^{5}}{5} - \frac{16 ( 16 + x ^{2} ) ^{3}}{3} : \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}}

= \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (16 + x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 + x^{2})^{\frac{3}{2}} + C

d er i v a t i v e 4 x (x^{2} - 9)

y^{^{''}} + 3 y^{^{'}} + 10 y = 0

s l o p e (- 6, - 6), (- 5, 0)

\frac{d}{d x} (x^{3} + 5 x^{2} - x^{- 2} + 8)

\frac{d}{d x} (sin^{7} (2 x))

What is the integral of x^3sqrt(16+x^2) ?
The integral of x^3sqrt(16+x^2) is 1/5 (16+x^2)^{5/2}-16/3 (16+x^2)^{3/2}+C