What is the integral of 9/((3x-4)^2) ?

The integral of 9/((3x-4)^2) is -3/(3x-4)+C

integral of 9/((3x-4)^2)

integral

\frac{9}{( 3 x - 4 ) ^{2}}

- \frac{3}{3 x - 4} + C

Solution steps

\int \frac{9}{( 3 x - 4 ) ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( 3 x - 4 ) ^{2}} d x

Apply u-substitution

= 9 \cdot \int \frac{1}{3 u ^{2}} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= 9 \cdot \frac{1}{3} (- \frac{1}{u})

Substitute back

u = 3 x - 4

= 9 \cdot \frac{1}{3} (- \frac{1}{3 x - 4})

Simplify

9 \cdot \frac{1}{3} (- \frac{1}{3 x - 4}) : - \frac{3}{3 x - 4}

= - \frac{3}{3 x - 4}

Add a constant to the solution

= - \frac{3}{3 x - 4} + C

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What is the integral of 9/((3x-4)^2) ?
The integral of 9/((3x-4)^2) is -3/(3x-4)+C