Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$, 
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$

$$\int  \frac { dx }{ 2x^{ 2 }+x+3 } =\int { \frac { dx }{ 2\left( { x }^{ 2 }+\frac { x }{ 2 } +\frac { 3 }{ 2 }  \right)  } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+\frac { x }{ 2 } +\frac { 1 }{ 16 } -\frac { 1 }{ 16 } +\frac { 3 }{ 2 }  }  }  } =$$ $$\frac { 1 }{ 2 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+\frac { 23 }{ 16 }  }  } =\\ =\frac { 1 }{ 2 } \int { \frac { dx }{ \frac { 23 }{ 16 } \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+1 \right]  }  } =\frac { 8 }{ 23 } \int { \frac { dx }{ \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+1 \right]  }  } =\frac { 8 }{ 23 } \frac { \sqrt { 23 }  }{ 4 } \int { \frac { d\left( \frac { 4 }{ \sqrt { 23 }  } x+\frac { 1 }{ \sqrt { 23 }  }  \right)  }{ { \left( \frac { 4 }{ \sqrt { 23 }  } x+\frac { 1 }{ \sqrt { 23 }  }  \right)  }^{ 2 }+1 }  } =\frac { 2 }{ \sqrt { 23 }  } \arctan { \left( \frac { 4x+1 }{ \sqrt { 23 }  }   \right) +C } $$
