How to solve $\int \frac{1}{x^2+4x+7} dx$? How to solve $\int \frac{1}{x^2+4x+7} dx$?
I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$
 A: A second step could be the change of variable
$$
x+2=\sqrt{3}u, \implies (x+2)^2+3=3u^2+3=3(u^2+1)
$$ giving
$$
\int \frac{1}{(x+2)^2+3} dx=\frac1{\sqrt{3}}\int \frac{1}{u^2+1} du
$$ which is now a standard integral.
A: With the substitution $u=\frac{x+2}{\sqrt{3}}$:
\begin{align*}\int \frac{1}{(x+2)^2+3} \mathrm{d}x &= \int \frac{1}{3(\frac{x+2}{\sqrt{3}})^2+3} \mathrm{d}x \\  &= \frac13\int \frac{1}{(\frac{x+2}{\sqrt{3}})^2+1} \mathrm{d}x \\ &= \frac{1}{3}\sqrt{3}\int \frac{1}{u^2+1} \mathrm{d}u  \\  &= \frac{1}{3}\sqrt{3} \arctan(u)+C \\ &= \frac{1}{3}\sqrt{3} \arctan\left(\frac{x+2}{\sqrt{3}}\right)+C \end{align*}
A: $$\int  \frac { 1 }{ (x+2)^{ 2 }+3 } dx=\int { \frac { dx }{ 3\left( 1+{ \left( \frac { x+2 }{ \sqrt { 3 }  }  \right)  }^{ 2 } \right)  }  } =\frac { 1 }{ \sqrt { 3 }  } \int { \frac { d\left( \frac { x+2 }{ \sqrt { 3 }  }  \right)  }{ 1+{ \left( \frac { x+2 }{ \sqrt { 3 }  }  \right)  }^{ 2 } }  } =\\=\frac { 1 }{ \sqrt { 3 }  } \arctan { { \left( \frac { x+2 }{ \sqrt { 3 }  }  \right)  } } +C$$
A: Continue on:
$$\int\frac{dx}{3+(x+2)^2}=\frac1{\sqrt3}\int\frac{\frac1{\sqrt3}dx}{1+\left(\frac{x+2}{\sqrt3}\right)^2}$$
and now you have an integral of the form 
$$\int\frac{f'}{1+f^2}dx$$
