What is the integral of 6/((2x-3)^2) ?

The integral of 6/((2x-3)^2) is -3/(2x-3)+C

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integral of 6/((2x-3)^2)

\int \frac{6}{( 2 x - 3 ) ^{2}} d x

- \frac{3}{2 x - 3} + C

Solution steps

\int \frac{6}{( 2 x - 3 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 2 x - 3 ) ^{2}} d x

Apply u-substitution

= 6 \cdot \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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= 6 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 x - 3

= 6 \cdot \frac{1}{2} (- \frac{1}{2 x - 3})

Simplify

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

= - \frac{3}{2 x - 3}

Add a constant to the solution

= - \frac{3}{2 x - 3} + C

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What is the integral of 6/((2x-3)^2) ?
The integral of 6/((2x-3)^2) is -3/(2x-3)+C