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

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

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

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

- \frac{1}{16 ( 8 y - 3 ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

Substitute back

u = 8 y - 3

= \frac{1}{8} (- \frac{1}{2 ( 8 y - 3 ) ^{2}})

Simplify

\frac{1}{8} (- \frac{1}{2 ( 8 y - 3 ) ^{2}}) : - \frac{1}{16 ( 8 y - 3 ) ^{2}}

= - \frac{1}{16 ( 8 y - 3 ) ^{2}}

Add a constant to the solution

= - \frac{1}{16 ( 8 y - 3 ) ^{2}} + C

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