Solve |2x+1|-1/2ln(x^2+x+1)+1/sqrt{3}arctan2x/1

$$| 2 x + 1 | - \frac { 1 } { 2 } \ln ( x ^ { 2 } + x + 1 ) + \frac { 1 } { \sqrt { 3 } } \arctan \frac { 2 x } { 1 }$$

Factor

$$\frac{-3\ln(x^{2}+x+1)+2\sqrt{3}\arctan(2x)+6|2x+1|}{6}$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\arctan(\frac{2x}{1}))$$

The square of \sqrt{3} is 3.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{3}\arctan(\frac{2x}{1}))$$

Anything divided by one gives itself.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}}{3}\arctan(2x))$$

Express \frac{\sqrt{3}}{3}\arctan(2x) as a single fraction.

$$factor(|2x+1|-\frac{1}{2}\log_{e}\left(x^{2}+x+1\right)+\frac{\sqrt{3}\arctan(2x)}{3})$$

Factor out \frac{1}{6}.

$$\frac{6|2x+1|-3\log_{e}\left(x^{2}+x+1\right)+2\sqrt{3}\arctan(2x)}{6}$$

Evaluate

$$\frac{-3\ln(x^{2}+x+1)+2\sqrt{3}\arctan(2x)+6|2x+1|}{6}$$