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$\quad$I was reviewing integration of rational functions, and all was going well until I saw this bit, in the end of the explanation: (translated from portuguese by me)

$ \qquad \qquad\text{with }x^2+bx + c \text{being a polynomial with } \alpha\pm \beta i\text{ as roots (no real roots),} $ $\qquad\qquad\text{ and using the previous variable change,} \text{ it is a good exercise to prove that:}$

$$\int \left( {D+Ex \over x^2+bx+c}\right)dx = \frac E2 \ln[(x-\alpha)^2+\beta^2] + {D+E\alpha\over \beta} \arctan\left({x-\alpha\over\beta}\right) + C $$

The variable change suggested was $x-\alpha = \beta t$

I can get close, but I always end up with a factor of $\frac1\beta$ that shouldn't be there. Here's what I've got:

$\quad$since $\alpha \pm \beta i$ are the roots of the polynomial, we can write $x^2+bx+c$ as $(x-\alpha)^2+\beta^2$.

$$\text{From the variable change, we get:}$$

$$x-\alpha=\beta t\qquad x=\beta t+\alpha\qquad t={x-\alpha\over\beta}$$

So:

$$\int \left( {D+Ex \over x^2+bx+c}\right)dx = \int \left( {D+Ex \over (x-\alpha)^2 +\beta^2} \right)dx$$

Performing the variable change:

$$(...)=\int \left( {D+E(\alpha +\beta t) \over \beta^2 t^2 + \beta^2}\right)dt = \int \left( {D+E\alpha +E\beta t) \over \beta^2 t^2 + \beta^2}\right)dt = \int \left( {E\beta t \over \beta^2 t^2 + \beta^2} + {D+E\alpha \over \beta^2 t^2 + \beta^2}\right)dt = \int \left( {E\beta t \over \beta^2 t^2 + \beta^2}\right)dt + \int\left({D+E\alpha \over \beta^2 t^2 + \beta^2}\right)dt$$

Let's solve these separately:

$$\int \left( {E\beta t \over \beta^2 t^2 + \beta^2}\right)dt = \int \left( {E\over 2\beta} {2\beta^2 t \over\beta^2t^2 + \beta^2}\right)dt$$

We now have a case of ${f'(t)\over f(t)}$, where $f(t)= \beta^2 t+\beta^2$ and $f'(t) = 2\beta^2 t$

$$(...)= {E\over 2\beta} \ln|\beta^2 t+\beta^2| + C_1$$ going from t back to x: $$(...)= \frac E2 \frac 1\beta \ln|\beta^2 {(x-\alpha)^2\over \beta^2} + \beta^2| + C_1= \frac E2 \frac 1\beta \ln|(x-\alpha)^2 + \beta^2|+ C_1$$

$\quad$Now to the second integral: $$\int \left({ D+E\alpha \over\beta^2t^2 + \beta^2}\right)dt = \int \left( (D+E\alpha) {1\over\beta^2t^2 + \beta^2}\right)dt = {D+E\alpha\over \beta^2}\int\left( {1\over t^2 +1}\right)dt$$

So now we have ${f'(t)\over f(t)^2+1}$, where $f(t) = t$ and, of course, $f'(t) = 1$:

$$(...) = {D+E\alpha\over \beta} \frac 1\beta \arctan(t) + C_2$$ going back to x: $$(...) = {D+E\alpha\over \beta} \frac 1\beta \arctan\left({x-\alpha\over\beta}\right) + C_2$$

So: $$\int \left( {D+Ex \over x^2+bx+c}\right)dx = \frac E2 \frac 1\beta \ln|(x-\alpha)^2 + \beta^2| + {D+E\alpha\over \beta} \frac 1\beta \arctan\left({x-\alpha\over\beta}\right) + C_1+ C_2 = \frac 1\beta \left[\frac E2 \ln|(x-\alpha)^2 + \beta^2| + {D+E\alpha\over \beta} \arctan\left({x-\alpha\over\beta}\right)\right] + C $$

Comparing this result with what we were trying to prove, it's clear the presence of $\frac1\beta$ (It could also be multiplying C, since C isn't a specific number). So... I've looked this over and over, and can't find a missing multiplication by $\beta$ or any other mistake that can explain this difference between my result and the right one. Any help?

Edit: Alright, this is a little fuzzy in my memory, so correct me if I'm wrong, but when changing from $\int dt$ to $\int dx$ I have to multiply the expression by $(\beta t + \alpha)' = \beta$, right? That would explain my error...

Sampaio
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2 Answers2

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You forgot to multiply by the derivative when you made the substitution.You need to multiply by $\dfrac{dx}{dt}(t)\color{grey}{=\beta}$ and your problems are over.

Theorem: Let $I,J$ be two non-trivial intervals, $f\colon I\to \Bbb R$ a function and $\varphi \colon J\to I$ a differentiable bijection.
If $f$ has an antiderivative (in $I$), then $$\int f(x)\, \mathrm dx = \int f(\varphi (t))\varphi '(t)\,\mathrm dt.$$

In your example $\varphi$ is the function $t\mapsto \beta t+\alpha.$

Git Gud
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  • Uhm, yes, I believe that is the solution. Since my memory seems to be having trouble kick-starting, when exactly do I multiply by beta? when changing from t to x or from x to t? (so, 1st substitution or when returning to original variable?) – Sampaio Sep 02 '13 at 17:13
  • @Sampaio When you make the first subtituion. Let me add some details to the answer. – Git Gud Sep 02 '13 at 17:14
  • Perfectly simple explanation, and it does, quite easily and beautifuly, solve my problem. A multiplication for beta will simplify with 1/beta and give me the right final result. Answer accepted – Sampaio Sep 02 '13 at 17:32
  • @Sampaio $\ddot \smile$ – Git Gud Sep 02 '13 at 17:33
  • Oh, and not to be a nitpicker, and not that it matters, but my close-to-OCD is making me tell you that you forgot a space in the first "to multiply". – Sampaio Sep 02 '13 at 17:34
  • @Sampaio Ahaha. Actually I'd rather you tell me about such typos, any typos for that matter. Thanks. – Git Gud Sep 02 '13 at 17:36
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You can always check an indefinite integration by differentiating. Try it with $\alpha = 0$, $\beta = 2$.

Robert Israel
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