What is the integral of [(2x-3)/(x^2+2x+5)] ?

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

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integral of [(2x-3)/(x^2+2x+5)]

\int [\frac{2 x - 3}{x ^{2} + 2 x + 5}] d x

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

Solution steps

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

Expand

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

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

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

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

Add similar elements:

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

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

Add a constant to the solution

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

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