Integration question (don't know what to substitute).

Question

How do you integrate the following? $$\int\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}dx$$ Please do not give the full answer, but just what to start with. Thanks.

Answer 1

Substitution would be a good start and there are several options, such as: $$t=\sqrt{\frac{1+x}{1-x}}$$ Squaring and solving for $x$: $$t^2=\frac{1+x}{1-x} \iff x = \frac{t^2-1}{t^2+1} \implies \mbox{d}x = \left(\frac{t^2-1}{t^2+1}\right)' \,\mbox{d}t $$ And you end up with a rational integrand.

Answer 2

Hint

Set $\;\displaystyle t=\biggl(\frac{1+x}{1-x}\biggr)^{\!\tfrac12}\iff x=\frac{t^2-1}{t^2+1}, \;t\ge 0$.

Answer 3

Hint :

If $\displaystyle I=\int \sqrt {\dfrac{1+x}{1-x}}dx $

put $x=-y$

$$2I =I +I=?$$

Answer 4

You may first solve the question a bit before substituting, like this:

(Multiplying the numerator and denominator by $\sqrt{1-x}$)

$$\sqrt{\frac {1+x} {1-x}} = \frac {\sqrt{1-x^2}} {1-x}$$

And then whenever you have something with like $1-x^2$ and $x$ in numerator or denominator along with other, the best and most obvious substitution mostly is $x=\sin\theta$ or $x=\cos\theta$, as you guess, we can get something really helpful in $1-x^2$ after this substitution.

Answer 5

Hint For $|x|<1$ $$ \sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{1+x}{1-x}\cdot\frac{1+x}{1+x}}=\frac{1+x}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}=\dfrac{d}{dx}\left[\sin^{-1}(x)-\sqrt{1-x^2}\right]. $$