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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int 2 (u^{2} - 1) d u

= - \int 2 (u^{2} - 1) 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} - 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

= - 2 (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= - 2 (\frac{u ^{3}}{3} - u)

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\int (x + x) d x

\int \frac{ln ( x )}{x ^{3}} d x

\frac{d u}{d u}

d v = cos (3 x) d x

\int (4^{u} + ln (5^{u})) d u

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