What is the integral of 1/((-3x+1)^7) ?

The integral of 1/((-3x+1)^7) is 1/(18(-3x+1)^6)+C

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integral of 1/((-3x+1)^7)

\int \frac{1}{( - 3 x + 1 ) ^{7}} d x

\frac{1}{18 ( - 3 x + 1 ) ^{6}} + C

Solution steps

\int \frac{1}{( - 3 x + 1 ) ^{7}} d x

Apply u-substitution

: \int - \frac{1}{3 u ^{7}} d u

= \int - \frac{1}{3 u ^{7}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int \frac{1}{u ^{7}} d u

Apply the Power Rule

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

= - \frac{1}{3} (- \frac{1}{6 u ^{6}})

Substitute back

u = - 3 x + 1

= - \frac{1}{3} (- \frac{1}{6 ( - 3 x + 1 ) ^{6}})

Simplify

- \frac{1}{3} (- \frac{1}{6 ( - 3 x + 1 ) ^{6}}) : \frac{1}{18 ( - 3 x + 1 ) ^{6}}

= \frac{1}{18 ( - 3 x + 1 ) ^{6}}

Add a constant to the solution

= \frac{1}{18 ( - 3 x + 1 ) ^{6}} + C

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What is the integral of 1/((-3x+1)^7) ?
The integral of 1/((-3x+1)^7) is 1/(18(-3x+1)^6)+C