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

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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 (sin (x) - 2 cos (x)) d x

\frac{2 s}{s ^{2} + 4}

y = ln (8 - 3 x)

\int e^{- u} u d u

x sin (\frac{8}{x})

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