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

The integral of (2x-3)/(x^2+x+5) is ln|4x^2+4x+20|-8/(sqrt(19))arctan(1/(sqrt(19))(2x+1))+C

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

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

ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1)) + C

Solution steps

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

Expand

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

\int \frac{2 x}{x ^{2} + x + 5} d x = ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19})

\int \frac{3}{x ^{2} + x + 5} d x = \frac{6}{19} arctan (\frac{2 x + 1}{19})

= ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19}) - \frac{6}{19} arctan (\frac{2 x + 1}{19})

Simplify

ln 4 x^{2} + 4 x + 20 - \frac{2}{19} arctan (\frac{2 x + 1}{19}) - \frac{6}{19} arctan (\frac{2 x + 1}{19}) : ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1))

= ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1))

Add a constant to the solution

= ln 4 x^{2} + 4 x + 20 - \frac{8}{19} arctan (\frac{1}{19} (2 x + 1)) + C

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What is the integral of (2x-3)/(x^2+x+5) ?
The integral of (2x-3)/(x^2+x+5) is ln|4x^2+4x+20|-8/(sqrt(19))arctan(1/(sqrt(19))(2x+1))+C