What is the integral of 1/((x+1)^5) ?

The integral of 1/((x+1)^5) is -1/(4(x+1)^4)+C

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integral of 1/((x+1)^5)

\int \frac{1}{( x + 1 ) ^{5}} d x

- \frac{1}{4 ( x + 1 ) ^{4}} + C

Solution steps

\int \frac{1}{( x + 1 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{u ^{5}} d u

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= - \frac{1}{4 u ^{4}}

Substitute back

u = x + 1

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

Add a constant to the solution

= - \frac{1}{4 ( x + 1 ) ^{4}} + C

x \to 0 lim (\frac{e ^{b x} - 1}{x})

x \to + 1 lim (\frac{x}{( x - 1 )})

x \to 0 lim (\frac{x - 6}{x - 2})

\frac{d ^{3}}{d x ^{3}} (1 + x)

\frac{d y}{d x} = 12

What is the integral of 1/((x+1)^5) ?
The integral of 1/((x+1)^5) is -1/(4(x+1)^4)+C