What is the integral of 5/((3x-4)^2) ?

The integral of 5/((3x-4)^2) is -5/(3(3x-4))+C

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integral of 5/((3x-4)^2)

\int \frac{5}{( 3 x - 4 ) ^{2}} d x

- \frac{5}{3 ( 3 x - 4 )} + C

Solution steps

\int \frac{5}{( 3 x - 4 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( 3 x - 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2}} d u

= 5 \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

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 5 \cdot \frac{1}{3} (- \frac{1}{u})

Substitute back

u = 3 x - 4

= 5 \cdot \frac{1}{3} (- \frac{1}{3 x - 4})

Simplify

5 \cdot \frac{1}{3} (- \frac{1}{3 x - 4}) : - \frac{5}{3 ( 3 x - 4 )}

= - \frac{5}{3 ( 3 x - 4 )}

Add a constant to the solution

= - \frac{5}{3 ( 3 x - 4 )} + C

3 - 2 cos (x), at x = \frac{π}{2}

\int \frac{1}{12} ln^{2} (3 x + 1) d x

\int \frac{1}{2 + 2 x ^{2}} d x

1 - \frac{1}{t}

What is the integral of 5/((3x-4)^2) ?
The integral of 5/((3x-4)^2) is -5/(3(3x-4))+C