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I was evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ and got an answer which, I think, is not correct as it is different from wolframalpha's answer.

Here's my work: $$\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1}\\& =\int \frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx + \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2}\\&= \sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C\tag{3}\end{align}$$

Steps:

$(1.)$ Rationalized the numerator.

$(2.)$ Applied linearity.

$(3.)$ The first integral is done by substituting $x^2 - (3/2)^2 = t$ and the second one is inverse hyperbolic cosine.

WolframAlpha shows this. I also tried differentiating both the answers, but still it's different from mine.

Utkarsh
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    It seems that WA is handling the case $4x^2 < 9$ and you are handling the case $4x^2 > 9$. Extended to the complex plane I guess you'd have the same answer. I haven't done the algebra to check but I'd guess that's the issue. – preferred_anon Feb 11 '23 at 17:00
  • @SineoftheTime $$\int \frac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1} \left(\frac{x}{a}\right)$$ – Utkarsh Feb 11 '23 at 17:03
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    To buttress @preferred_anon's observation, note that for the second integral in (3) the WA result is this. It's an ugly-looking answer, but the point of interest here is that this antiderivative vanishes at the origin...but $1/\sqrt{x^2-(3/2)^2}$ is imaginary at the origin, so this is outside the domain of the real antiderivative with domain $x\in [3/2,\infty)$. (This is one of those cases where indefinite integration can be rather deceiving.) – Semiclassical Feb 11 '23 at 17:05
  • Aah! This one is a little complicated. – Utkarsh Feb 11 '23 at 17:08
  • I changed my mind, I think WA is just wrong. The $\sin^{-1}$ term is only real for $1 < x < 3/2$, outside the domain of your integrand. – preferred_anon Feb 11 '23 at 17:33
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    My own way of avoiding these issues when they arise is to not work with indefinite integrals as such. Instead I prefer integrals that have a fixed basepoint, e.g., $F(x)=\int_{3/2}^x \sqrt{\frac{2t+3}{2t-3}},dt$ where $x\geq 3/2$. – Semiclassical Feb 11 '23 at 17:50
  • @Utkarsh Yep, I misread the text. I thought it was $x^2+(3/2)^2$ – Sine of the Time Feb 11 '23 at 17:56

1 Answers1

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\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1p}\\& =\int \frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx + \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2p}\\&= \sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C\tag{3p}\end{align}

  1. Your work is correct, but on precisely the interval $(1.5,\infty).$ This is because

    • step $(1\mathrm p)$ assumes that $(2x+3)\ge0.$
    • step $(3\mathrm p)$ assumes that $x\ge0.$

  2. For the other piece $(-\infty,-1.5]$ of the integrand's domain: \begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int -\frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1n}\\& =\int -\frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx - \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2n}\\&= -\sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(-\frac{2x}{3}\right) + D.\tag{3n}\end{align}

  3. Then the final answer is the piecewise function combining the above two pieces, noting that they have independent arbitrary constants $C$ and $D.$

This Desmos graph verifies that the red curve (the given integrand) comprises precisely the blue curve (the derivative function of the positive piece $(3\mathrm p)$) and the green curve (the derivative function of the negative piece $(3\mathrm n)$).

ryang
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  • @Semiclassical The piecewise function exhibits the domain as two separate pieces!, whereas giving the simplified one-line single-piece one-integration-constant answer conveys no such info, so I don't understand your argument. (I argued the same thing in the first paragraph of the first link in the Answer. :) – ryang Feb 11 '23 at 18:15
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    Thinking about it more, I yield the point. – Semiclassical Feb 11 '23 at 19:27
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    Did you check the Wolfram Alpha answer OP posted? I think it's just wrong. – preferred_anon Feb 11 '23 at 19:30
  • @preferred_anon I just did, and updated the linked Desmos graph to include the Wolfram results: (1) it gives the correct plot of the antiderivatives (see graphs #4,5), but (2) its symbolic answer is undefined throughout $\mathbb R$ due to $\sqrt{\frac{2x+3}{2x-3}}$ and $\left(\sqrt{9-4x^{2}}+6\arcsin\left(\frac{\sqrt{3-2x}}{\sqrt{6}}\right)\right)$ overlapping at exactly the one point for which its denominator is undefined (see graphs #6-10). – ryang Feb 12 '23 at 07:16