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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:

enter image description here

I don't see how they are equal, but apparently they are, if anyone could illuminate it.

Anonymous Computer
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bobby
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4 Answers4

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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: $$\frac {x-\alpha}{x-\beta}\in \mathbb{R}^{+}$$

Lucian
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Graham Hesketh
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$$\int\frac{dx}{x^2+1}=\arctan x=\frac1{2i}\cdot\ln\bigg[\frac{x-i}{x+i}\bigg]$$

$$\int\frac{dx}{x^2-1}=\arctan ix=\frac12\cdot\ln\bigg[\frac{x-1}{x+1}\bigg]$$

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 !

Lucian
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  • Cool, can you turn the exact logarithm in my post into an arctan without complex numbers? – bobby May 28 '14 at 00:18
  • @bobby: $\alpha$ and $\beta$ are either both real, or both complex, depending on whether $\Delta=b^2-4ac$ is positive or negative. Now, symbolically, both formulas are correct, but practically, the arctangent is useful in the latter case, and the logarithm in the former. – Lucian May 28 '14 at 00:23
  • I don't know how to get the arctan solution without memorizing a bunch of simplifications, but I can remember the logarithm method of solution - can we get the arctan form from this? i.e. can we derive the arctan form of the solution, in my post, starting from the logarithm form of solution in my post? – bobby May 28 '14 at 00:25
  • @bobby: Sure! Use the fact that $\alpha$ and $\beta$ are both of the form $A\pm Bi$, in conjunction with the fact that $i\cdot\ln\dfrac{U-Vi}{U+Vi}=\arctan\dfrac{UV}2$ , where $U=x-A$, and $V=B$. – Lucian May 28 '14 at 00:56
  • Setting $\alpha = \frac{-b}{2} +\frac{\sqrt{(b/a)^2-4(c/a)}}{2}$ I get $\frac{1}{a}\frac{-2}{\sqrt{(b/a)^2-4(c/a)}} \arctan (\frac{(x+\frac{b}{2a}\sqrt{(b/a)^2-4(c/a)}}{4})$, which doesn't look equal to the expression in my original post, but it's an answer (I hope is okay) – bobby May 28 '14 at 01:36
  • @bobby: I gave you the wrong expression! It's $\arctan\dfrac{2UV}{U^2-V^2}$ , if $U>V>0$. – Lucian May 28 '14 at 02:10
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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.

David
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    What does one do in general? My edit to the post seems to say the formulas do work in general, what's going on? – bobby May 27 '14 at 23:59
  • I don't think so, read the last line I added after your edit. – David May 28 '14 at 00:00
  • Assuming arbitrary real values for a, b and c, can I use the logarithm method of solution, then convert it into something adapted to whatever $b^2-4ac$ dictates? If not, what should I do? – bobby May 28 '14 at 00:01
  • Yeah, I'm ultimately just trying to find the simplest, quickest way to integrate $\frac{1}{(ax^2+bx+c)^n}$ http://math.stackexchange.com/questions/811772/indefinite-integral-of-frac1ax2bxcn without any memorization or simplifications so if they'll help please use them, I think allowing complex roots into the integral avoids long-winded recursion relations as we do not need to integrate the quadratic we can reduce it to linear terms, then get rid of them at the end – bobby May 28 '14 at 00:06
  • Do it as a logarithm if the quadratic has real roots, otherwise an inverse tangent. – David May 28 '14 at 00:09
  • If it has real roots, great. If it has equal roots, I can do it with a u-substitution, great. If it has complex roots, it looks very similar to the arctan forms on this page http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms with i's and that page says it's formally equivalent to an arctan, why does the method not work? It looks like it should? – bobby May 28 '14 at 00:16
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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.

GEdgar
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