What is the integral of ((x^2+9))/(x^2+2x+3) ?

The integral of ((x^2+9))/(x^2+2x+3) is x-ln|x^2+2x+3|+4sqrt(2)arctan(1/(sqrt(2))(x+1))+C

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integral of ((x^2+9))/(x^2+2x+3)

\int \frac{( x ^{2} + 9 )}{x ^{2} + 2 x + 3} d x

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

Solution steps

\int \frac{x ^{2} + 9}{x ^{2} + 2 x + 3} d x

Long division

\frac{x ^{2} + 9}{x ^{2} + 2 x + 3} : 1 + \frac{- 2 x + 6}{x ^{2} + 2 x + 3}

= \int 1 + \frac{- 2 x + 6}{x ^{2} + 2 x + 3} d x

Apply the Sum Rule:

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

= \int 1 d x + \int \frac{- 2 x + 6}{x ^{2} + 2 x + 3} d x

\int 1 d x = x

\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))

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

Add a constant to the solution

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

y^{^{''}} + y^{^{'}} = e^{x}

\int \frac{1}{x ^{- 1}} d x

\frac{d y}{d t} = y (10 - y)

\frac{\partial}{\partial x} (cos (θ))

y = - x - 1, x = - 1, x = 2, y = 0

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