What is the integral of (3x+1)/(x^2+2x+5) ?

The integral of (3x+1)/(x^2+2x+5) is 3(1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1)))+1/2 arctan((x+1)/2)+C

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integral of (3x+1)/(x^2+2x+5)

\int \frac{3 x + 1}{x ^{2} + 2 x + 5} d x

3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + \frac{1}{2} arctan (\frac{x + 1}{2}) + C

Solution steps

\int \frac{3 x + 1}{x ^{2} + 2 x + 5} d x

Expand

\frac{3 x + 1}{x ^{2} + 2 x + 5} : \frac{3 x}{x ^{2} + 2 x + 5} + \frac{1}{x ^{2} + 2 x + 5}

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

\int \frac{3 x}{x ^{2} + 2 x + 5} d x = 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)))

\int \frac{1}{x ^{2} + 2 x + 5} d x = \frac{1}{2} arctan (\frac{x + 1}{2})

= 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + \frac{1}{2} arctan (\frac{x + 1}{2})

Add a constant to the solution

= 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + \frac{1}{2} arctan (\frac{x + 1}{2}) + C

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What is the integral of (3x+1)/(x^2+2x+5) ?
The integral of (3x+1)/(x^2+2x+5) is 3(1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1)))+1/2 arctan((x+1)/2)+C