How to evaluate this integral? $\displaystyle I_1=\int\frac{\left(\frac{a}{y}-\frac{a}{b}\right)^{1/2}}{1-\frac{a}{y}}\mathrm {d}y$ Here $a$ and $b$ are real constants and $y$ real variable.
It is solvable by a method of integration by substitution, where new variable $t$ is introduced by some trigonometric combination, e.g. $y=b\cos^2t$. However the solution is not as expected. Can someone suggest a better new variable or another method?
The solution obtained via $y=b\cos^2t$ is $I_1=-\frac{1}{2}\sqrt{ab}\left\{2\left(1-\frac{a}{b}\right)\arccos\left(\frac{y}{b}\right)+4\sqrt{\frac{a}{b}\left(\frac{a}{b}-1\right)}\arctan\left(\frac{\frac{b}{y}-1}{1-\frac{b}{a}}\right)^{1/2}-2\sqrt{\frac{y}{b}\left(1-\frac{y}{b}\right)}\right\}$
The solution should be composed of functions $\arctan$, $\log$ and square roots.
It turned out that ther integral $I_1$ is not exactly what I should solve. (It is related to integral $I_2$ by addition of a number of complicated terms.)
The following integral is the problem: $I_2=-\left(1-\frac{a}{b}\right)^{1/2}\int_{b}^{x}\left(1-\frac{a}{y}\right)^{-1}\left(\frac{a}{y}-\frac{a}{b}\right)^{-1/2}dy\;,$ where $0
Result from WolframAlpha is not returned.
The proposed solution has the form $I_2=\left(\frac{b}{a}-1\right)^{1/2}b\sqrt{\frac{b}{x}-1}+a\left(\frac{b}{a}-1\right)^{1/2}\left(2+\frac{b}{a}\right)\arctan\left(\sqrt{\frac{b}{x}-1}\right)\sqrt{\frac{r_s}{x}}+a\log\left|\frac{\sqrt{\frac{b}{x}-1}+x^2}{\sqrt{\frac{b}{x}-1}-x^2}\right|$
The question is whether this proposed solution is correct and how to derive it?
Again by introducing a new variable $y=b\cos^2t$, I came to this solution of indefinite integral in $I_2$: $\left(-\left(1-\frac{a}{b}\right)^{1/2}\right)\left(-\frac{\sqrt{ab}}{2}\right)\left\{\left(2-\frac{a}{b}\right)\arccos\left(\frac{y}{b}\right)^{1/2}-2\left[\frac{y}{b}\left(\frac{y}{b}-1\right)\right]^{1/2}+4\left[\frac{a}{b}\left(\frac{a}{b}-1\right)\right]^{1/2}\arctan\left(\frac{\frac{b}{y}-1}{1-\frac{b}{a}}\right)^{1/2}\right\}$
Here is the result of indefinite integral in $I_2$ from WolframAlpha.
Could this lead to the proposed solution which contains $\log$ and $\arctan$ with a different argument?