Integral of $\int \frac{\sqrt{a^2-x^2}}{x} \, dx$ and the result of WolframAlpha

Question

In solving the integral $\int \frac{\sqrt{a^2-x^2}}{x} \, dx$ I have found out by hand thet the result is $$|a|\sqrt{1-\bigg(\frac{x}{a}\bigg)^2}+\frac{|a|}{2}\ln\Bigg({\frac{\bigg|1-\sqrt{1-\bigg(\frac{x}{a}\bigg)^2}\bigg|}{\bigg|1+\sqrt{1-\bigg(\frac{x}{a}\bigg)^2}\bigg|}}\Bigg)+c=\sqrt{a^2-x^2}-|a|\ln{\Bigg(\frac{|a|\big(\sqrt{1-\frac{x^2}{a^2}}+1\big)}{|x|}}\Bigg)+c$$ But in order to check my result I have write the integral on wolfram and it tells me that is: $$\sqrt{(a^2 - x^2)} - a \tanh^{-1}{(\frac{\sqrt{a^2 - x^2}}{a}}) \,\,\,\,(*)$$ Wolfram Alpha

I don't know if really my result and $(*)$ are equivalent. Can you help me?

Answer 1

Since $\operatorname{artanh}u=\tfrac12\ln\tfrac{1+u}{1-u}$, Wolfram Alpha's formula is $\sqrt{a^2-x^2}-\tfrac12a\ln\tfrac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}+c$, in agreement with your calculation.

Answer 2

The equivalence arises from an elementary hyperbolic trigonometric identity. Recall $$y = \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}},$$ hence solving this equation for $x$ yields $$x = \frac{1}{2} \log \frac{1+y}{1-y} = \tanh^{-1} y.$$ This is actually what you wrote in your first step when simplifying your antiderivative, with the choice $$y = \frac{\sqrt{a^2 - x^2}}{a}.$$