Evaluate $\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}dx$

Question

Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$

Answer 1

HINT:

Setting $x=\cos2y$

$$I=\int\dfrac{2-3x}{2+3x}\sqrt{\dfrac{1+x}{1-x}}dx=-2\int\dfrac{2-3\cos2y}{2+3\cos2y}\cdot\cot y\sin2y\ dy$$

$$=-2\int\dfrac{2(1+\tan^2y)-3(1-\tan^2y)}{2(1+\tan^2y)+3(1-\tan^2y)}\cos^2y\ dy$$

$$=-2\int\dfrac{5\tan^2y-1}{(5-\tan^2y)(1+\tan^2y)^2}\sec^2y\ dy$$

Set $\tan y=u$ to find $$I=-2\int\dfrac{5u^2-1}{(5-u^2)(1+u^2)^2}du$$

Now use Partial Fraction Decomposition,

$$\dfrac{5v-1}{(5-v)(1+v)^2}=\dfrac A{5-v}+\dfrac B{1+v}+\dfrac C{(1+v)^2}$$

Answer 2

The substitution $t = \sqrt{\frac{1 + x}{1 - x}}$, that is, $$x = \frac{t^2 - 1}{t^2 + 1}, \qquad dx = \frac{4 t \,dt}{(t^2 + 1)^2},$$ rationalizes the integrand.