What is the integral of (12u)/(sqrt(u(u-8))) ?

The integral of (12u)/(sqrt(u(u-8))) is 12(sqrt(u^2-8u)+4ln(1/4 |sqrt(u^2-8u)+u-4|))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (12u)/(sqrt(u(u-8)))

\int \frac{12 u}{u ( u - 8 )} d u

12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4)) + C

Solution steps

\int \frac{12 u}{u ( u - 8 )} d u

Take the constant out:

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

= 12 \cdot \int \frac{u}{u ( u - 8 )} d u

Complete the square

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

= 12 \cdot \int \frac{u}{( u - 4 ) ^{2} - 16} d u

Apply u-substitution

: \int \frac{v + 4}{v ^{2} - 16} d v

= 12 \cdot \int \frac{v + 4}{v ^{2} - 16} d v

Expand

\frac{v + 4}{v ^{2} - 16} : \frac{v}{v ^{2} - 16} + \frac{4}{v ^{2} - 16}

= 12 \cdot \int \frac{v}{v ^{2} - 16} + \frac{4}{v ^{2} - 16} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 12 (\int \frac{v}{v ^{2} - 16} d v + \int \frac{4}{v ^{2} - 16} d v)

\int \frac{v}{v ^{2} - 16} d v = v^{2} - 16

\int \frac{4}{v ^{2} - 16} d v = 4 ln (\frac{1}{4} v^{2} - 16 + v)

= 12 (v^{2} - 16 + 4 ln (\frac{1}{4} v^{2} - 16 + v))

Substitute back

v = u - 4

= 12 ((u - 4)^{2} - 16 + 4 ln (\frac{1}{4} (u - 4)^{2} - 16 + u - 4))

Simplify

12 ((u - 4)^{2} - 16 + 4 ln (\frac{1}{4} (u - 4)^{2} - 16 + u - 4)) : 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4))

= 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4))

Add a constant to the solution

= 12 (u^{2} - 8 u + 4 ln (\frac{1}{4} u^{2} - 8 u + u - 4)) + C

k = 0 \sum \infty (k + 1) x^{k}

5 x^{2} e^{3 x}

x \to - \frac{π}{4} + lim (- sec (2 x))

x \cdot \frac{d y}{d x} = y + x^{2} - y^{2}

x (x + 5 y) (\frac{d y}{d x}) + (y (3 x + 5 y)) = 0

What is the integral of (12u)/(sqrt(u(u-8))) ?
The integral of (12u)/(sqrt(u(u-8))) is 12(sqrt(u^2-8u)+4ln(1/4 |sqrt(u^2-8u)+u-4|))+C