what is $\int{\sqrt{\frac{x}{x-c}}} dx$

Let $x=c\cosh^2y\;\Rightarrow\;dx=2c\cosh y\sinh y\ dy$, then $$\eqalign { \int\sqrt{\frac{x}{x-c}}\ dx&\stackrel{\color{red}{(1)}}=2c\int\cosh^2y\ dy\\ &=2c\int\frac{1+\cosh2y}{2}\ dy\\ &=c\left(y+\frac12\sinh 2y\right)+K_1\\ &\stackrel{\color{red}{(2)}}=c\left(\text{arccosh}\ \sqrt{\frac xc}+\sinh y\cosh y\right)+K_1\\ &=c\ln\left(\sqrt{\frac xc}+\sqrt{\frac{x-c}{c}}\right)+c\sqrt{\frac{x-c}{c}}\cdot\sqrt{\frac{x}{c}}+K_1\\ &\stackrel{\color{red}{(3)}}=c\ln\left(\sqrt x+\sqrt{x-c}\right)-\frac c2\ln c+\sqrt{x(x-c)}+K_1\\ &\stackrel{\color{red}{(4)}}=\color{blue}{c\ln\left(\sqrt x+\sqrt{x-c}\right)+\sqrt{x(x-c)}+K_2}. }$$ The final result is equal to output of Wolfram Alpha.

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Notes:

$\color{red}{(1)}\ \ \cosh^2y-\sinh^2y=1$

$\color{red}{(2)}\ \ \text{arccosh}\ x=\ln\left(x+\sqrt{x^2-1}\right)\quad;\quad x\ge1$

$\color{red}{(3)}\ \ \ln\dfrac ab=\ln a-\ln b\ \text{and}\ \ln a^b=b\ln a$

$\color{red}{(4)}\ \ K_2=K_1-\dfrac c2\ln c$

Let $u^2 = \dfrac{x}{x-c} \to x = u^2(x-c) = u^2x - u^2c \to u^2c = (u^2-1)x \to x = \dfrac{u^2c}{u^2-1} = c + \dfrac{c}{u^2 -1}$. Thus: $dx = \dfrac{-2cu}{(u^2-1)^2}du$, and we have:

$I = -2c\displaystyle \int \dfrac{u^2}{(u^2-1)^2}du$, and you can use fraction decomposition to continue.