What is the integral of (6x)/((9x^2-4)^3) ?

The integral of (6x)/((9x^2-4)^3) is -1/(6(9x^2-4)^2)+C

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integral of (6x)/((9x^2-4)^3)

\int \frac{6 x}{( 9 x ^{2} - 4 ) ^{3}} d x

- \frac{1}{6 ( 9 x ^{2} - 4 ) ^{2}} + C

Solution steps

\int \frac{6 x}{( 9 x ^{2} - 4 ) ^{3}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 9 x ^{2} - 4 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{18 u ^{3}} d u

= 6 \cdot \int \frac{1}{18 u ^{3}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{18} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 6 \cdot \frac{1}{18} (- \frac{1}{2 u ^{2}})

Substitute back

u = 9 x^{2} - 4

= 6 \cdot \frac{1}{18} (- \frac{1}{2 ( 9 x ^{2} - 4 ) ^{2}})

Simplify

6 \cdot \frac{1}{18} (- \frac{1}{2 ( 9 x ^{2} - 4 ) ^{2}}) : - \frac{1}{6 ( 9 x ^{2} - 4 ) ^{2}}

= - \frac{1}{6 ( 9 x ^{2} - 4 ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( 9 x ^{2} - 4 ) ^{2}} + C

\frac{\partial}{\partial x} x y \cdot e^{(\frac{( - 5 x ^{2} + 5 y ^{2} )}{2} + 1)}

\frac{d y}{d x} = y^{2} (6 - 2 x)

\frac{\partial}{\partial x} (ln (\frac{x}{k}))

\frac{d}{d x} ((2 - x) (y - 2) (3 - x - y))

\frac{d}{d x} ((3 x y + 7)^{2})

What is the integral of (6x)/((9x^2-4)^3) ?
The integral of (6x)/((9x^2-4)^3) is -1/(6(9x^2-4)^2)+C