What is the integral of (20)/((5x-9)^2) ?

The integral of (20)/((5x-9)^2) is -4/(5x-9)+C

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integral of (20)/((5x-9)^2)

\int \frac{20}{( 5 x - 9 ) ^{2}} d x

- \frac{4}{5 x - 9} + C

Solution steps

\int \frac{20}{( 5 x - 9 ) ^{2}} d x

Take the constant out:

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

= 20 \cdot \int \frac{1}{( 5 x - 9 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{5 u ^{2}} d u

= 20 \cdot \int \frac{1}{5 u ^{2}} d u

Take the constant out:

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

= 20 \cdot \frac{1}{5} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 20 \cdot \frac{1}{5} (- \frac{1}{u})

Substitute back

u = 5 x - 9

= 20 \cdot \frac{1}{5} (- \frac{1}{5 x - 9})

Simplify

20 \cdot \frac{1}{5} (- \frac{1}{5 x - 9}) : - \frac{4}{5 x - 9}

= - \frac{4}{5 x - 9}

Add a constant to the solution

= - \frac{4}{5 x - 9} + C

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What is the integral of (20)/((5x-9)^2) ?
The integral of (20)/((5x-9)^2) is -4/(5x-9)+C