What is the integral of (x^2)/(sqrt(x+4)) ?

The integral of (x^2)/(sqrt(x+4)) is 2(1/5 (x+4)^{5/2}-8/3 (x+4)^{3/2}+16sqrt(x+4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(x+4))

\int \frac{x ^{2}}{x + 4} d x

2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}} + 16 x + 4) + C

Solution steps

\int \frac{x ^{2}}{x + 4} d x

Apply u-substitution

: \int 2 (u^{2} - 4)^{2} d u

= \int 2 (u^{2} - 4)^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (u^{2} - 4)^{2} d u

Expand

(u^{2} - 4)^{2} : u^{4} - 8 u^{2} + 16

= 2 \cdot \int u^{4} - 8 u^{2} + 16 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

= 2 (\int u^{4} d u - \int 8 u^{2} d u + \int 16 d u)

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

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

\int 16 d u = 16 u

= 2 (\frac{u ^{5}}{5} - \frac{8 u ^{3}}{3} + 16 u)

Substitute back

u = x + 4

= 2 (\frac{( x + 4 ) ^{5}}{5} - \frac{8 ( x + 4 ) ^{3}}{3} + 16 x + 4)

Simplify

2 (\frac{( x + 4 ) ^{5}}{5} - \frac{8 ( x + 4 ) ^{3}}{3} + 16 x + 4) : 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}} + 16 x + 4)

= 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}} + 16 x + 4)

Add a constant to the solution

= 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{8}{3} (x + 4)^{\frac{3}{2}} + 16 x + 4) + C

\frac{d}{d x} (3 sin^{5} (x))

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

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

\frac{d}{d t} (v (t))

\frac{d}{d x} (\frac{f ( x )}{g ( x )})

What is the integral of (x^2)/(sqrt(x+4)) ?
The integral of (x^2)/(sqrt(x+4)) is 2(1/5 (x+4)^{5/2}-8/3 (x+4)^{3/2}+16sqrt(x+4))+C