What is the integral of 1/((3x-1)^4) ?

The integral of 1/((3x-1)^4) is -1/(9(3x-1)^3)+C

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integral of 1/((3x-1)^4)

\int \frac{1}{( 3 x - 1 ) ^{4}} d x

- \frac{1}{9 ( 3 x - 1 ) ^{3}} + C

Solution steps

\int \frac{1}{( 3 x - 1 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{4}} d u

= \int \frac{1}{3 u ^{4}} 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 ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = 3 x - 1

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

Simplify

\frac{1}{3} (- \frac{1}{3 ( 3 x - 1 ) ^{3}}) : - \frac{1}{9 ( 3 x - 1 ) ^{3}}

= - \frac{1}{9 ( 3 x - 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{9 ( 3 x - 1 ) ^{3}} + C

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

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\int - sec^{3} (x) sin (x) d x

\int_{7}^{e^{4}} \frac{1}{x} d x

7 x - x^{2}, 2 x, 0, 5

What is the integral of 1/((3x-1)^4) ?
The integral of 1/((3x-1)^4) is -1/(9(3x-1)^3)+C