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

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

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

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

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

Solution steps

\int (1 + x)^{- 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 = 1 + x

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

Add a constant to the solution

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

\frac{d}{d x} (- 3 x + 5)

\int_{0}^{1} x \cdot arctan (x^{2}) d x

y = 4 - x

7 x + 2

\frac{d}{d x} (ln (\frac{a - x}{a + x}))

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