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I'm in the chapter of trigonometric substitution for integrating different functions. I'm having a bit of trouble even starting this homework question: $$\int \frac{(x^2+3x+4)\,dx} {\sqrt{x^2-4x}}$$

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A start: We have $x^2-4x=(x-2)^2-4$. If you want to use a trigonometric substitution, the natural one is $x-2=2\sec t$.

We unfortunately end up needing to integrate powers of $\sec t$, a messy business. Nicer is the hyperbolic function substitution $x-2=2\cosh t$.

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Also note that $x^{2}+3x+4 = (x-2)^{2} + 7x = 4\sec^{2}(t) + 7 \cdot (2\sec(t)+2)$

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Is this a comment to someone else's answer? It does not appear to be an answer. – robjohn Jun 13 '13 at 17:47
@robjohn, thank you, I will edit. – Brady Trainor Jun 14 '13 at 14:09

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Make the substitution suggested by completing the square, $\sqrt{x^2-4x}=\sqrt{(x-2)^2-4}$, that is, $y=x-2$. Then, by the easily constructed triangle, we have

\begin{align} \frac{y}{2} & = \sec\theta\\ \frac{\sqrt{y^2-4}}{2} & = \tan\theta\\ x^2+3x+4 & =(y+2)^2+3(y+2)+4\\ & = y^2+7y+14\\ \frac{dy}{2} & = \sec\theta\tan\theta\ d\theta. \end{align} The integral becomes

\begin{align} \int \frac{(x^2+3x+4)} {\sqrt{x^2-4x}}dx & = \int\frac{y^2+7y+14}{\sqrt{y^2-4}}dy\\ & = \int\frac{4\sec^2\theta+14\sec\theta+14}{2\tan\theta}2\sec\theta\tan\theta\ d\theta\\ & = \int4\sec^3\theta+14\sec^2\theta+14\sec\theta\ d\theta. \end{align}

Andre Nicolas suggests in another response that the powers of secant are unsavory, and that a hyperbolic trig sub would be more apt. (Claude Leibovici explores this.)

(I vaguely recall methods for the $\sec^3\theta$ integrand in Stewart's Calculus. The $\sec\theta$ is in most trig integral lists, and perhaps the $\sec^2\theta$ is amenable to some fortuitous trig identity.)

Here are some resources on those:

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Unfortunately $\sqrt{\vphantom{()^2}x^2-4x}\ne\sqrt{(x-2)^2+4}$ – robjohn Jun 13 '13 at 17:45
@robjohn, thank you, I will edit. – Brady Trainor Jun 14 '13 at 14:09

In order to make a proper substitution in integral calculus, the function that you are substituting must have a unique inverse function. However, there is such a case where the the derivative is present and you can make what I refer to as a "virtual substitution". This is not exactly the case here, we have to do other algebraic manipulations. Trigonometric functions such as sine, cosine and their variants have infinitely many inverse functions; inverse trigonometric functions ( i.e. arcsine, arccosine, etc...) have a unique inverse function, thus are fine. For example, if I made the substitution $y = \sin x$ ( where $-1≤y≤1$) , then $ x = (-1)^n \cdot \arcsin y + n\pi$ ($n \in \mathbb Z$): this does not work, without bound. If anyone disagrees with my statement, please prove that the substitution is proper. Also, in my opinion, turning a rational/algebraic function into a transcendental function is ridiculous. There are very elementary ways to approach this integral; a good book to read on many of these methods is Calculus Made Easy by Silvanus P. Thompson.

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May be I was too fast even if the answer is almost correct.

First make $x = 2 (\cosh(y) + 1)$. The integrand then becomes

$(\pm) 2 (8 + 7 \cosh(y) + \cosh(2 y))$

So, the integral is $(\pm) (16 y + 14 \sinh(y) + \sinh(2 y))$

If you replace $y$, you arrive to my formula (the $\pm$ was missing)

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$%1/2 Sqrt[(-4 + x) x] (12 + x + (64 Log[Sqrt[-4 + x] + Sqrt[x]])/(Sqrt[-4 + x] Sqrt[x]))$ $$ \frac12\sqrt{(-4+x)x}\left(12+x+\frac{64\log(\sqrt{-4+x}+\sqrt{x})}{\sqrt{-4+x}\sqrt{x}}\right) $$

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The question was about how to do the problem, not what the answer is. – robjohn Jun 13 '13 at 17:43

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