What is the integral of (x+4)sqrt(8x+x^2) ?

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

\int (x + 4) 8 x + x^{2} d x

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

Solution steps

\int (x + 4) 8 x + x^{2} d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = 8 x + x^{2}

= \frac{( 8 x + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 8 x + x ^{2} ) ^{3}}{3} : \frac{1}{3} (8 x + x^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d}{d x} (r^{2} - 7 x^{2})

\frac{d}{d x} (2 sec^{2} (x) tan (x))

x \to 0 lim (\frac{5}{x - 1})

sin (2 t + 3)

y = x, x \geq 0

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