What is the integral of (3x+6)/((2x^2+8x+3)^2) ?

The integral of (3x+6)/((2x^2+8x+3)^2) is -3/(4(2x^2+8x+3))+C

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integral of (3x+6)/((2x^2+8x+3)^2)

\int \frac{3 x + 6}{( 2 x ^{2} + 8 x + 3 ) ^{2}} d x

- \frac{3}{4 ( 2 x ^{2} + 8 x + 3 )} + C

Solution steps

\int \frac{3 x + 6}{( 2 x ^{2} + 8 x + 3 ) ^{2}} d x

Apply u-substitution

: \int \frac{27 u}{( 2 u ^{2} - 45 ) ^{2}} d u

= \int \frac{27 u}{( 2 u ^{2} - 45 ) ^{2}} d u

Take the constant out:

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

= 27 \cdot \int \frac{u}{( 2 u ^{2} - 45 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{4 v ^{2}} d v

= 27 \cdot \int \frac{1}{4 v ^{2}} d v

Take the constant out:

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

= 27 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 27 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 27 \cdot \frac{1}{4} (- \frac{1}{2 ( 3 x + 6 ) ^{2} - 45})

Simplify

27 \cdot \frac{1}{4} (- \frac{1}{2 ( 3 x + 6 ) ^{2} - 45}) : - \frac{3}{4 ( 2 x ^{2} + 8 x + 3 )}

= - \frac{3}{4 ( 2 x ^{2} + 8 x + 3 )}

Add a constant to the solution

= - \frac{3}{4 ( 2 x ^{2} + 8 x + 3 )} + C

y = \frac{8}{x - 2}, at x = 6

\frac{d}{d x} (26 e^{x^{2}} x)

\frac{\partial}{\partial y} (5 x sin (y - z))

\frac{d}{d x} ((7 x^{2} - 12 x) e^{x})

y^{^{'}} - e^{5 x + y} = 0

What is the integral of (3x+6)/((2x^2+8x+3)^2) ?
The integral of (3x+6)/((2x^2+8x+3)^2) is -3/(4(2x^2+8x+3))+C