What is the integral of 1/((8x-1)^3) ?

The integral of 1/((8x-1)^3) is -1/(16(8x-1)^2)+C

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integral of 1/((8x-1)^3)

\int \frac{1}{( 8 x - 1 ) ^{3}} d x

- \frac{1}{16 ( 8 x - 1 ) ^{2}} + C

Solution steps

\int \frac{1}{( 8 x - 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{8 u ^{3}} d u

= \int \frac{1}{8 u ^{3}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{8} (- \frac{1}{2 u ^{2}})

Substitute back

u = 8 x - 1

= \frac{1}{8} (- \frac{1}{2 ( 8 x - 1 ) ^{2}})

Simplify

\frac{1}{8} (- \frac{1}{2 ( 8 x - 1 ) ^{2}}) : - \frac{1}{16 ( 8 x - 1 ) ^{2}}

= - \frac{1}{16 ( 8 x - 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{16 ( 8 x - 1 ) ^{2}} + C

\int \frac{2 x ^{2} - 1}{( 4 x - 1 ) ( x ^{2} + 1 )} d x

f (x) = a^{2}

\frac{x ^{2} - 49}{x - 7}

f (x) = 4 x - 3 x^{2}

(y - 2) d y + x d x = 0

What is the integral of 1/((8x-1)^3) ?
The integral of 1/((8x-1)^3) is -1/(16(8x-1)^2)+C