# Calculate the integral $\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$ [duplicate]

I have to calculate the integral $$\int \frac{2-3x}{2+3x} \sqrt{\frac{1+x}{1-x}}dx$$. I tried the following substitutions: $$x \rightarrow \frac{1+t}{1-t}, x \rightarrow \frac{1-t}{1+t}, x \rightarrow \frac{t^{2}+1}{t^{2}-1}$$ but with no good result. Also I observed the symmetry in the integral, by $$x\rightarrow\frac{x}{3}$$. However I am unable to end it.

## marked as duplicate by Zacky, Eevee Trainer, Alex Provost, Lord Shark the Unknown, dantopaMar 8 at 5:58

• How about $x=\cos2t$ – lab bhattacharjee Mar 7 at 18:06
I would substitute $$t=\sqrt{\frac{1+x}{1-x}}$$ so $$x=\frac{t^2-1}{t^2+1}$$ and $$dx=\frac{4t}{(t^2+1)^2}dt$$ and our integral will be $$\int -4\,{\frac {{t}^{2} \left( {t}^{2}-5 \right) }{ \left( 5\,{t}^{2}-1 \right) \left( {t}^{2}+1 \right) ^{2}}} dt$$
Hint The standard substitution for integrals like this, where the integrand includes $$\sqrt{\frac{1 + x}{1 - x}}$$, is simply $$u = \sqrt{\frac{1 + x}{1 - x}}$$. Rearranging and differentiating gives $$x = \frac{u^2 - 1}{u^2 + 1}, \qquad dx = \frac{4 u \,du}{(u^2 + 1)^2}.$$ So in our case, where the integrand is a product of $$\sqrt{\frac{1 + x}{1 - x}}$$ and a rational function of $$x$$, the substitution produces a rational function: $$\int \frac{2 - 3 x}{2 + 3 x} \sqrt{\frac{1 + x}{1 - x}} \,dx = -\frac{4}{5} \int \frac{(u^2 - 5) u^2\,du}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} .$$ As usual we apply the method of partial fractions, that is write the integrand as $$\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{A u + B}{(u^2 + 1)^2} + \frac{C u + D}{u^2 + 1} + \frac{E}{u - \frac{1}{\sqrt{5}}} + \frac{F}{u + \frac{1}{\sqrt{5}}} .$$ With six parameters to solve for, this is a priori a computationally involved problem. But the left-hand side is an even function of $$u$$, so the right-hand side must be, too, which immediately gives $$A = C = 0$$, $$F = -E$$, so we only need to solve for three parameters: $$\boxed{\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{B}{(u^2 + 1)^2} + \frac{D}{u^2 + 1} + \frac{E}{u - \frac{1}{\sqrt{5}}} - \frac{E}{u + \frac{1}{\sqrt{5}}}} .$$
(Instead of decomposing fully, we can also decompose $$\frac{(u^2 - 5) u^2}{(u^2 + 1)^2 (u^2 - \frac{1}{5})} = \frac{B}{(u^2 + 1)^2} + \frac{D u}{u^2 + 1} + \frac{E'}{u^2 - \frac{1}{5}} ,$$ and then use the elementary integral $$\int \frac{du}{u^2 - a^2} = \frac{1}{2 a }\log \left\vert\frac{u - a}{u + a}\right\vert + C$$.)
Alternatively, as Lab pointed out in the comments, the form $$\sqrt{\frac{1 + x}{1 - x}}$$ suggests substitutions using various trigonometric identities, for example, $$\cot \theta = \pm \sqrt\frac{1 + \cos 2 \theta}{1 - \cos 2\theta} .$$