What is the integral of 1/((2y-3)^2) ?

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

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integral of 1/((2y-3)^2)

\int \frac{1}{( 2 y - 3 ) ^{2}} d y

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

Solution steps

\int \frac{1}{( 2 y - 3 ) ^{2}} d y

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 2 y - 3

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

Simplify

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

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

Add a constant to the solution

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

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