What is the integral of (1+x)^{-4} ?

The integral of (1+x)^{-4} is -1/(3(1+x)^3)+C

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integral of (1+x)^{-4}

\int (1 + x)^{- 4} d x

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

Solution steps

\int (1 + x)^{- 4} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 1 + x ) ^{4}} d x

Apply u-substitution

= \int \frac{1}{u ^{4}} d u

Apply the Power Rule

= - \frac{1}{3 u ^{3}}

Substitute back

u = 1 + x

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

Add a constant to the solution

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

\frac{7 d u}{d t} = u^{2}

4 y^{^{''}} + 8 y = 0

x \to - 1 lim (sin (π x))

\frac{d}{d x} (2 x y^{3})

a re a x^{3} - 10 x^{2} + 16 x, - x^{3} + 10 x^{2} - 16 x

What is the integral of (1+x)^{-4} ?
The integral of (1+x)^{-4} is -1/(3(1+x)^3)+C