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

The integral of (-2x+6)/(x^2+2x+3) is -ln|x^2+2x+3|+4sqrt(2)arctan(1/(sqrt(2))(x+1))+C

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

\int \frac{- 2 x + 6}{x ^{2} + 2 x + 3} d x

- ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)) + C

Solution steps

\int \frac{- 2 x + 6}{x ^{2} + 2 x + 3} d x

Expand

\frac{- 2 x + 6}{x ^{2} + 2 x + 3} : - \frac{2 x}{x ^{2} + 2 x + 3} + \frac{6}{x ^{2} + 2 x + 3}

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

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

\int \frac{6}{x ^{2} + 2 x + 3} d x = 32 arctan (\frac{1}{2} (x + 1))

= - (ln x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 1))) + 32 arctan (\frac{1}{2} (x + 1))

Simplify

= - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)) + C

y^{^{''}} + 8 y^{^{'}} + 15 y = 2 e^{- 5 x}

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 25})

\int e^{t x - x} d x

\frac{\partial}{\partial y} (7 x + 8 y^{2})

\int e^{x} cos (y) + yz d x

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