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

The integral of 5/((3x-1)^5) is -5/(12(3x-1)^4)+C

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

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

- \frac{5}{12 ( 3 x - 1 ) ^{4}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u ^{5}} d u

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

Apply the Power Rule

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

= 5 \cdot \frac{1}{3} (- \frac{1}{4 u ^{4}})

Substitute back

u = 3 x - 1

= 5 \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x - 1 ) ^{4}})

Simplify

5 \cdot \frac{1}{3} (- \frac{1}{4 ( 3 x - 1 ) ^{4}}) : - \frac{5}{12 ( 3 x - 1 ) ^{4}}

= - \frac{5}{12 ( 3 x - 1 ) ^{4}}

Add a constant to the solution

= - \frac{5}{12 ( 3 x - 1 ) ^{4}} + C

y = tan (arcsin (x))

\frac{\partial}{\partial x} (4 x^{2} y^{3} - 4)

y^{^{'}} = 0.2 ln (\frac{3000}{y}) y, y (0) = 1000

\int \frac{x + 2}{x ^{2} + x + 1} d x

\frac{d}{d x} (π^{2})

What is the integral of 5/((3x-1)^5) ?
The integral of 5/((3x-1)^5) is -5/(12(3x-1)^4)+C