# Integration of a square root of a reciprocal function

I've been trying to solve this integration for an investigation but have no idea where to start: $$\int\sqrt{\frac{a+x}{b-x}}dx$$ I've tried integrating by parts, but seem to end up with an even more complicated integral. I also tried to move the $x$ to the denominator: $$\int\sqrt{\frac{a+b}{b-x}-1}dx$$ But I don't think I've ever solved an integral like this before.

with $$t=\sqrt{\frac{a+x}{b+x}}$$ we get $$x=\frac{t^2b-a}{1+t^2}$$ and $$dx=\frac{2t(a+b)}{(t^2+1)^2}dt$$ and our integral is given by now as $$\int\frac{2t^2(a+b)}{(t^2+1)^2}dt$$

• From here we can substitute $\tan(x)$ for $t$, correct? – John Lou Oct 9 '17 at 18:13
• yes this is possible – Dr. Sonnhard Graubner Oct 9 '17 at 18:14

Assuming $b > -a$, for $-a \leqslant x < b$ the integral can be found by "rationalising the numerator" first.

So \begin{align*} \int \sqrt{\frac{a + x}{b - x}} \, dx &= \int \sqrt{\frac{a + x}{b - x}} \cdot \frac{\sqrt{a + x}}{\sqrt{a + x}} \, dx\\ &= \int \frac{a + x}{\sqrt{(b - x)(a + x)}} \, dx\\ &= \int \frac{a + x}{\sqrt{ab - (a - b)x - x^2}} \, dx \end{align*} Now rewriting the numerator as the derivative of the denominator we have \begin{align*} \int \sqrt{\frac{a + x}{b - x}} \, dx &= -\frac{1}{2} \int \frac{(-a + b - 2x) - a -b}{\sqrt{ab - (a - b)x - x^2}} \, dx\\ &= - \frac{1}{2} \int \frac{-(a - b) - 2x}{\sqrt{ab - (a - b) x - x^2}} \, dx + \frac{a + b}{2} \int \frac{dx}{\sqrt{ab - (a - b) x - x^2}} \, dx\\ &= I_1 + I_2 \end{align*} The first of these integrals can be found using a substitution of $u = ab - (a - b)x - x^2$. The result is $$I_1 = -\sqrt{(b - x)(a + x)} + C_1.$$ The second integral can be found by first completing the square. As $$ab - (a - b)x - x^2 = \left (\frac{a + b}{2} \right )^2 - \left (x + \frac{a - b}{2} \right )^2,$$ we have $$I_2 = \frac{a + b}{2} \int \frac{dx}{\sqrt{\left (\frac{a + b}{2} \right )^2 - \left (x + \frac{a - b}{2} \right )^2}} \, dx = \frac{a + b}{2} \sin^{-1} \left ( \frac{2x + a - b}{a + b} \right ) + C_2.$$ The result then follows by adding the two together.

Hint:

If $x\ne -a,$ we need $$(x+a)(x-b)<0$$

Assuming $b>-a,$ we need $$-a<x<b$$

$\iff p-a<p+x<p+b$

If we set $p+b=-(p-a),2p=a-b\implies\dfrac{a-b}2-a<\dfrac{a-b}2+x<\dfrac{a-b}2+b$

Set $\dfrac{a-b}2+x=\dfrac{a+b}2\cdot\cos2t$

$\iff x=b\cos^2t-a\sin^2t$

$x+a=?,b-x=?$