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

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

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integral of (4x+4)/((x^2+2x+3)^2)

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

- \frac{2}{x ^{2} + 2 x + 3} + C

Solution steps

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

Expand

\frac{4 x + 4}{( x ^{2} + 2 x + 3 ) ^{2}} : \frac{4 x}{( x ^{2} + 2 x + 3 ) ^{2}} + \frac{4}{( x ^{2} + 2 x + 3 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{4 x}{( x ^{2} + 2 x + 3 ) ^{2}} d x + \int \frac{4}{( x ^{2} + 2 x + 3 ) ^{2}} d x

\int \frac{4 x}{( x ^{2} + 2 x + 3 ) ^{2}} d x = - \frac{2}{x ^{2} + 2 x + 3} - \frac{1}{2} arctan (\frac{1}{2} (x + 1)) - \frac{1}{2 2} sin (2 arctan (\frac{1}{2} (x + 1)))

\int \frac{4}{( x ^{2} + 2 x + 3 ) ^{2}} d x = \frac{1}{2} (arctan (\frac{1}{2} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x + 1))))

= - \frac{2}{x ^{2} + 2 x + 3} - \frac{1}{2} arctan (\frac{1}{2} (x + 1)) - \frac{1}{2 2} sin (2 arctan (\frac{1}{2} (x + 1))) + \frac{1}{2} (arctan (\frac{1}{2} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x + 1))))

Simplify

- \frac{2}{x ^{2} + 2 x + 3} - \frac{1}{2} arctan (\frac{1}{2} (x + 1)) - \frac{1}{2 2} sin (2 arctan (\frac{1}{2} (x + 1))) + \frac{1}{2} (arctan (\frac{1}{2} (x + 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x + 1)))) : - \frac{2}{x ^{2} + 2 x + 3}

= - \frac{2}{x ^{2} + 2 x + 3}

Add a constant to the solution

= - \frac{2}{x ^{2} + 2 x + 3} + C

f (x) = 6 x, at x = 6

\int \frac{t}{t ^{2} - 1} d t

f (x) = \frac{4}{x}, at 9 \frac{4}{9}

\int \frac{1}{2 x ^{2} + 2 x + 5} d x

\frac{\partial}{\partial x} (x^{\frac{3}{4}} \cdot y^{- 4})

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