What is the integral of (2t+2)/(t^2+t+1) ?

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

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integral of (2t+2)/(t^2+t+1)

\int \frac{2 t + 2}{t ^{2} + t + 1} d t

ln 4 t^{2} + 4 t + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 t + 1)) + C

Solution steps

\int \frac{2 t + 2}{t ^{2} + t + 1} d t

Expand

\frac{2 t + 2}{t ^{2} + t + 1} : \frac{2 t}{t ^{2} + t + 1} + \frac{2}{t ^{2} + t + 1}

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 t}{t ^{2} + t + 1} d t + \int \frac{2}{t ^{2} + t + 1} d t

\int \frac{2 t}{t ^{2} + t + 1} d t = ln 4 t^{2} + 4 t + 4 - \frac{2}{3} arctan (\frac{2 t + 1}{3})

\int \frac{2}{t ^{2} + t + 1} d t = \frac{4}{3} arctan (\frac{2 t + 1}{3})

= ln 4 t^{2} + 4 t + 4 - \frac{2}{3} arctan (\frac{2 t + 1}{3}) + \frac{4}{3} arctan (\frac{2 t + 1}{3})

Simplify

ln 4 t^{2} + 4 t + 4 - \frac{2}{3} arctan (\frac{2 t + 1}{3}) + \frac{4}{3} arctan (\frac{2 t + 1}{3}) : ln 4 t^{2} + 4 t + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 t + 1))

= ln 4 t^{2} + 4 t + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 t + 1))

Add a constant to the solution

= ln 4 t^{2} + 4 t + 4 + \frac{2}{3} arctan (\frac{1}{3} (2 t + 1)) + C

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