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

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

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integral of 1/((x+4)^5)

\int \frac{1}{( x + 4 ) ^{5}} d x

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

Solution steps

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

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

Add a constant to the solution

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

\int x x + 7 d x

x \to - 2 + lim (x)

\frac{d}{d x} (\frac{sin ( 2 x + 1 )}{2 x + 1})

\int - x^{- 4} d x

\int t^{4} d t

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