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

The integral of ((x-1))/(x^2+2x+5) is 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 ((x-1))/(x^2+2x+5)

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

\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{x - 1}{x ^{2} + 2 x + 5} d x

Expand

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

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

= \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

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