What is the integral of 5/((3x-1)^7) ?

The integral of 5/((3x-1)^7) is -5/(18(3x-1)^6)+C

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integral of 5/((3x-1)^7)

\int \frac{5}{( 3 x - 1 ) ^{7}} d x

- \frac{5}{18 ( 3 x - 1 ) ^{6}} + C

Solution steps

\int \frac{5}{( 3 x - 1 ) ^{7}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( 3 x - 1 ) ^{7}} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{7}} d u

Apply the Power Rule

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

= 5 \cdot \frac{1}{3} (- \frac{1}{6 u ^{6}})

Substitute back

u = 3 x - 1

= 5 \cdot \frac{1}{3} (- \frac{1}{6 ( 3 x - 1 ) ^{6}})

Simplify

5 \cdot \frac{1}{3} (- \frac{1}{6 ( 3 x - 1 ) ^{6}}) : - \frac{5}{18 ( 3 x - 1 ) ^{6}}

= - \frac{5}{18 ( 3 x - 1 ) ^{6}}

Add a constant to the solution

= - \frac{5}{18 ( 3 x - 1 ) ^{6}} + C

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

\frac{4 s}{s ^{2} + 6 s + 9}

x \to 0 lim ((1 + 3 x)^{\frac{7}{x}})

5 sin (6 t)

\int t e^{- 4 t} d t

What is the integral of 5/((3x-1)^7) ?
The integral of 5/((3x-1)^7) is -5/(18(3x-1)^6)+C