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: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
 A: $$\int  \frac { dx }{ 2x^{ 2 }+x+3 } =\int { \frac { dx }{ 2\left( { x }^{ 2 }+\frac { x }{ 2 } +\frac { 3 }{ 2 }  \right)  } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+\frac { x }{ 2 } +\frac { 1 }{ 16 } -\frac { 1 }{ 16 } +\frac { 3 }{ 2 }  }  }  } =$$ $$\frac { 1 }{ 2 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+\frac { 23 }{ 16 }  }  } =\\ =\frac { 1 }{ 2 } \int { \frac { dx }{ \frac { 23 }{ 16 } \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+1 \right]  }  } =\frac { 8 }{ 23 } \int { \frac { dx }{ \left[ \frac { 16 }{ 23 } { \left( x+\frac { 1 }{ 4 }  \right)  }^{ 2 }+1 \right]  }  } =\frac { 8 }{ 23 } \frac { \sqrt { 23 }  }{ 4 } \int { \frac { d\left( \frac { 4 }{ \sqrt { 23 }  } x+\frac { 1 }{ \sqrt { 23 }  }  \right)  }{ { \left( \frac { 4 }{ \sqrt { 23 }  } x+\frac { 1 }{ \sqrt { 23 }  }  \right)  }^{ 2 }+1 }  } =\frac { 2 }{ \sqrt { 23 }  } \arctan { \left( \frac { 4x+1 }{ \sqrt { 23 }  }   \right) +C } $$
A: HINT:
the second integral $\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx$
You should be able to use some substitutions to turn it into $C\int\frac{1}{u^2+1}du$
A: When you have a second degree polynomial in the denominator with $\Delta <0$, you can proceed like this: find the solutions, in this case $\frac{-1}{4} \pm \frac{i\sqrt {23}}{4}$, and then write the polynomial in the usual way
$$2\Big(x+\frac{1}{4} + \frac{i\sqrt {23}}{4}\Big)\Big(x+\frac{1}{4} - \frac{i\sqrt {23}}{4}\Big) $$
then you can use the fact that $(a-b)(a+b)=a^2-b^2$ to deduce
$$2x^2+x+3=2\Big(x+\frac{1}{4} \Big)^2+\frac{23}{8}.$$
Then apply a substitution to get $\frac{1}{1+y^2}$.
A: $$\int \frac{x}{2x^2+x+3}dx$$
$$=\frac{1}{4} \int \frac{4x+1}{2x^2+x+3} dx- \frac{1}{4} \int \frac{1}{2x^2+x+3}$$
I presume you know how to solve the first integral [Do so using the sub: $u=2x^2+x+3$].
Complete the square of the second and substitute $m=\sqrt{2}x+\frac{1}{2\sqrt{2}}$to get:
$$\frac{-1}{4}\int\frac{1}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}dx=\frac{-1}{4}\int\frac{1}{\sqrt{2}}\frac{1}{(m^2+\frac{23}{8})}dm$$
Factor out the 23/8 and the constants, to get $$\frac{-\sqrt{2}}{23}\int\frac{1}{\frac{8m^2}{23}+1}dm$$
Substitute $q=2\sqrt{\frac{2m}{23}}$, solve the integral (it should be easy knowing the derivative of arctan(q)). Substitute back everything.
You should eventually get the below as the answer:
$$\frac{1}{4}\ln(2x^2+x+3)-\frac{\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)}{2\sqrt{23}}+C$$
A: $$\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx= \frac{2}{\sqrt{23}}\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)+c$$
A: $$\int \frac{1}{2x^2+x+3}\, dx=\frac{1}{\sqrt{2}}\int \frac{d\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}\, $$
Let $\sqrt{2}x+\frac{1}{2\sqrt{2}}=\sqrt{\frac{23}{8}}\tan u$.
$$=\frac{1}{\sqrt{2}}\int \frac{\sqrt{\frac{23}{8}}\frac{1}{\cos^2 u}}{\frac{23}{8}(\tan^2 u+1)}\, du=\frac{2}{\sqrt{23}}\int  du=\frac{2}{\sqrt{23}}u+C$$
