Evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ 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.
 A: 
\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}


*

*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 $x\ge0.$



*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}


*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 $(3\mathrm n)$).
