What is the integral of (x+1)^{-3} ?

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

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integral of (x+1)^{-3}

\int (x + 1)^{- 3} d x

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

Solution steps

\int (x + 1)^{- 3} d x

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{2 u ^{2}}

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int \frac{x ^{2}}{x ^{3} - 2} d x

\frac{d y}{d x} = \frac{( 1 + x )}{x y ^{2}}, y (1) = 2

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

c f (x)

x \to 0 - lim (5 + \frac{2}{x})

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