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}

integral

(x + 1)^{- 3}

- \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

\frac{1}{a + x ^{2}}

((e)^{x})^{2}

cos^{2} (t) sin (t)

((sin (x))^{2}) ((cos^{2} (x))^{2})

\frac{1}{36 + x ^{2}}

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