What is the integral of (-8x)/((x+1)-sqrt(x+1)) ?

The integral of (-8x)/((x+1)-sqrt(x+1)) is -16(1/2 (x+1)+sqrt(x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (-8x)/((x+1)-sqrt(x+1))

\int \frac{- 8 x}{( x + 1 ) - x + 1} d x

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

Solution steps

\int \frac{- 8 x}{x + 1 - x + 1} d x

Take the constant out:

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

= - 8 \cdot \int \frac{x}{x + 1 - x + 1} d x

Apply u-substitution

= - 8 \cdot \int 2 (u + 1) d u

Take the constant out:

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

= - 8 \cdot 2 \cdot \int u + 1 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

= - 8 \cdot 2 (\int u d u + \int 1 d u)

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

\int 1 d u = u

= - 8 \cdot 2 (\frac{u ^{2}}{2} + u)

Substitute back

u = x + 1

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

Simplify

- 8 \cdot 2 (\frac{( x + 1 ) ^{2}}{2} + x + 1) : - 16 (\frac{1}{2} (x + 1) + x + 1)

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

Add a constant to the solution

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

What is the integral of (-8x)/((x+1)-sqrt(x+1)) ?
The integral of (-8x)/((x+1)-sqrt(x+1)) is -16(1/2 (x+1)+sqrt(x+1))+C