Arctan equalling a logarithm? Is
$$\int \frac{dx}{ax^2+bx+c}=\frac{1}{a}\int{dx}{(x-\alpha)(x-\beta)}=\frac{1}{a} \int \left( \frac{ \frac{1}{\alpha-\beta} }{x-\alpha} - \frac{\frac{1}{\alpha-\beta}}{x-\beta}\right)=\frac{1}{a} \frac{1}{\alpha-\beta} \ln\left|\frac{x-\alpha}{x-\beta}\right|$$
equal to 
$$\int \! \frac{1}{ a{x}^{2}+bx+c
 }{dx}=2\,\arctan \left( {
\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {\frac {1}{\sqrt {4\,c
a-{b}^{2}}}}$$
All I see is this
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms
edit:

I don't see how they are equal, but apparently they are, if anyone could illuminate it.
 A: $$\int\frac{dx}{x^2+1}=\arctan x=\frac1{2i}\cdot\ln\bigg[\frac{x-i}{x+i}\bigg]$$
$$\int\frac{dx}{x^2-1}=\arctan ix=\frac12\cdot\ln\bigg[\frac{x-1}{x+1}\bigg]$$
If your question is why the two forms are symbolically equivalent, the answer lies in the fact that trigonometric functions are related to the circle, whose equation is $x^2+y^2=r^2$, whereas the natural logarithm is related to the hyperbola, whose equation is $x^2-y^2=r^2$, which, after a rotation of $45^\circ$ and a bit of scaling, becomes $y=\dfrac1x$ , whose primitive is $\ln x$. Hope this helps !
A: Lemma: for $x\in \mathbb{R}^{+}$
$$\log(x)=2\,\mathrm{arctanh}\!\left(\frac{-1+x}{x+1}\right) \tag{1}$$
proof:
let $x=e^{2u},\,u\in \mathbb{R}$,
$$\begin{aligned}
\log(e^{2u})&=2\,\mathrm{arctanh}\!\left(\frac{-1+e^{2u}}{e^{2u}+1}\right)\\
2u&=2\,\mathrm{arctanh}\!\left(\frac{-1+e^{2u}}{e^{2u}+1}\right)=2\,\mathrm{arctanh}\!\left(\frac{-e^{-u}+e^{u}}{e^{u}+e^{-u}}\right)\\
2u&=2\,\mathrm{arctanh}\!\left(\frac{\sinh{(u)}}{\cosh{(u)}}\right)=2\,\mathrm{arctanh}\!\left(\tanh{(u)}\right)=2u\\
\end{aligned} \tag{2}$$ 
Now define $\alpha$ and $\beta$ via:
$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
$$\alpha=-{\frac {b}{2a}}+{\frac {\sqrt{{b}^{2}-4\,ca
}}{2a}},\quad\beta=-{\frac {b}{2a}}-{\frac {\sqrt{{b}^{2}-4\,ca}}{
2a}} \tag{3}$$
from $(3)$ it follows that:
$$
\begin{aligned}
a \left( \alpha+\beta \right) &=-b\\
a \left( \alpha-\beta \right) &=\sqrt {{b}^{2}-4\,ca}
\end{aligned} \tag{4}$$
we then note that, from:
$$\sqrt {4\,ca-{b}^{2}}=i\sqrt {{b}^{2}-4\,ca},\quad\arctan(ir)=i\,\mathrm{arctanh{(r)}}$$
and $(4)$, that:
$$
\begin{aligned}
2\,\arctan \left( {\frac {2\,ax+b}{\sqrt {4\,ca-{b}^{2}}}} \right) {
\frac {1}{\sqrt {4\,ca-{b}^{2}}}}&=\frac{2}{a\left( \alpha-\beta
 \right)}\,\mathrm{arctanh} \left( {\frac {-2\,
x+\alpha+\beta}{\alpha-\beta}} \right)\\
&=\frac{2}{a\left( \alpha-\beta
 \right)}\,\mathrm{arctanh} \left( {\frac {-1+{\frac {x-\alpha
}{x-\beta}}}{-1+{\frac {x-\alpha}{x-\beta}}}} \right)\\
&=\frac{2}{a\left( \alpha-\beta
 \right)}\,\log \left(\frac {x-\alpha
}{x-\beta}\right)
\end{aligned}
$$
where the last line follows from $(1)$ under the assumption that:
$$\frac {x-\alpha}{x-\beta}\in \mathbb{R}^{+}$$
A: Assuming that you want to do your integrals in terms of real numbers, . . . 


*

*if $b^2-4ac$ is positive then the second formula does not apply as you have the square root of a negative number;  

*if $b^2-4ac$ is negative the first formula does not apply since $\alpha,\beta$ are not real;  

*if $b^2-4ac$ is zero then neither formula is quite right.


In fact, if you carefully read the image you posted, this is exactly what it says.
A: Another way to look at it.  The tangent can be written in terms of complex exponentials.  Use this to solve (a quadratic equation) to write the arctangent in terms of complex logarithms.
