What is the integral of 1/((x-4)^7) ?

The integral of 1/((x-4)^7) is -1/(6(x-4)^6)+C

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integral of 1/((x-4)^7)

\int \frac{1}{( x - 4 ) ^{7}} d x

- \frac{1}{6 ( x - 4 ) ^{6}} + C

Solution steps

\int \frac{1}{( x - 4 ) ^{7}} d x

Apply u-substitution

: \int \frac{1}{u ^{7}} d u

= \int \frac{1}{u ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= - \frac{1}{6 u ^{6}}

Substitute back

u = x - 4

= - \frac{1}{6 ( x - 4 ) ^{6}}

Add a constant to the solution

= - \frac{1}{6 ( x - 4 ) ^{6}} + C

\int \frac{x + 11}{x ^{2} + 4 x + 8} d x

n = 1 \sum \infty \frac{1}{4 n + 2}

7 y ln (x) - x y^{^{'}} = 0

\frac{d}{d x} (\frac{x}{4 - x ^{2}})

g (u) = 3 u + 5 u

What is the integral of 1/((x-4)^7) ?
The integral of 1/((x-4)^7) is -1/(6(x-4)^6)+C