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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = 2 x - 3

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

Simplify

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

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

Add a constant to the solution

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

t an g e n t f (t) = 3 t - t^{2}, (0, 0)

\frac{d}{d x} (ln (\frac{1 + sin ( x )}{1 - sin ( x )}))

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

\int x^{2} + 4 x + 40 d x

d er i v a t i v e f (t) = 3 t sin (π t)

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