How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?

Question

How can I evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$$

This is taken from a definite integral where $x$ varies from $0$ to $1$.

My attempt:

Multiplied by conjugate $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}\\=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$ $$=\int \frac{\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$ $$=\int \frac{\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$

1. If I use $x=\sin^2\theta$ $$\int \frac{\left(\sqrt{1+\sin^2\theta}+\cos\theta\right)}{2\sin^2\theta}\sin2\theta\ d\theta\\ =\int \left(\sqrt{1+\sin^2\theta}+\cos\theta\right)\cot\theta d\theta$$
2. If I use $x=\tan^2\theta$ $$\int \frac{\left(\sec\theta-\sqrt{1-\tan^2\theta}\right)}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta\\ =\int \frac{\left(\sec\theta-\sqrt{1-\tan^2\theta}\right)}{\sin\theta\cos\theta} d\theta$$

Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.

Thanks

Answer 1

You can split the integral into two parts $$\int \frac{(\sqrt{1+x}+\sqrt{1-x})}{2x} \, dx=\frac{1}{2}\left[\int \frac{\sqrt{1+x}}{x}\,dx+\int \frac{\sqrt{1-x}}{x}\,dx\right].$$ Solve these separately as follows: \begin{align*} \int \frac{\sqrt{1+x}}{x}\,dx & =\int \frac{t^2}{(t^2-1)}\,dt && (\text{ let } 1+x=t^2) \\ & =\int \frac{t^2-1+1}{(t^2-1)}\,dt\\ & =\int 1 \, dt+\int \frac{1}{(t^2-1)}\,dt\\ & =t+\frac{1}{2}\left[\int \frac{1}{(t-1)}\,dt-\int \frac{1}{(t+1)}\,dt\right]\\ &=t+\ln\frac{|t-1|}{|t+1|}+c\\ &=\sqrt{1+x}+\ln\frac{|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|}+c\\ \end{align*} Observe that the second part is pretty much the same. If you use $x=-u$, then $$\int \frac{\sqrt{1-x}}{x}\, dx=\int \frac{\sqrt{1+u}}{u}\, du.$$ So you can write the answer without any further computation. $$\int \frac{\sqrt{1-x}}{x}\, dx=\sqrt{1\color{red}{-x}}+\ln\frac{|\sqrt{1\color{red}{-x}}-1|}{|\sqrt{1\color{red}{-x}}+1|}+c$$

Answer 2

$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$ $$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$ Let $x=\sin\theta\implies dx=\cos\theta d\theta$ $$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$ $$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \left(\because \cos\theta=1-2\sin^2\frac{\theta}{2}\right)$$ $$=\int \frac{\left(1-2\sin^2\frac{\theta}{2}\right)d\theta}{2\sin\frac{\theta}{2}}$$ $$=\int \left(\frac12\csc\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\ d\theta$$ $$=\ln \left|\tan\frac{\theta}{4}\right|+2\cos\frac{\theta}{2}+C$$

Answer 3

$\begin{aligned} I &:= \int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{2 x} d x =\frac{1}{2}\left[ \underbrace{\int \frac{\sqrt{1+x}}{x}d x}_{J}+\underbrace{\int \frac{\sqrt{1-x}}{x} d x}_{K}\right] \end{aligned}$ $$ \begin{aligned} J &=\int \frac{\sqrt{1+x}}{x} d x \\ &=\int \frac{1+x}{x \sqrt{1+x}} d x \\ &=\int \frac{1+x}{x} d (\sqrt{1+x}) \\ &=\int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\ &=\int \frac{d (\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+\sqrt{1+x} \\ &=\frac{1}{2} \ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\sqrt{1+x}+c_{1} \end{aligned} $$$$ K \stackrel{x\mapsto -x}{=} \sqrt{1-x}+\frac{1}{2} \ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+c_{2} $$ Now we can conclude that $$ I=\sqrt{1-x}+\sqrt{1+x}+\frac{1}{2} \ln \left| \frac{(\sqrt{1+x}-1)(\sqrt{1-x}-1)}{(\sqrt{1+x}+1)(\sqrt{1-x}+1)}\right|+C $$

Answer 4

Substitute $t= \sqrt{1+x}+\sqrt{1-x} $. Then \begin{align} &(\sqrt{1+x}-\sqrt{1-x})dx=(2-t^2 )dt\\ &(\sqrt{1+x}-\sqrt{1-x})^2=4-t^2 \end{align} and \begin{align} \int \frac{1}{\sqrt{1+x}-\sqrt{1-x}}\ dx =\int \frac{2-t^2}{4-t^2} dt= t-\tanh^{-1}\frac t2+C\\ \end{align}

Answer 5

With the change of variable $x=\sin 2t$, we have

$$\sqrt{1+x}-\sqrt{1-x}=\sqrt{\cos^2t+2\cos t\sin t+\sin^2t}-\sqrt{\cos^2t-2\cos t\sin t+\sin^2t}=2\sin t.$$

Then

$$\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int\frac{2\cos2t}{2\sin t}dt=\int\left(\frac1{\sin t}-2\sin t\right)dt \\=\text{arcoth}(\cos t)+2\cos t+C.$$

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From the biquadratic equation

$$4\cos^2t\,(1-\cos^2t)=x^2$$ you draw $\cos t$ as a function of $x$.

Answer 6

Let $I=\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$

By Rationalizing,

$I=\frac{1}{2}[\int\frac{\sqrt{1+x}} {{x}}dx + \int\frac{\sqrt{1-x}} {{x}}dx ]$

This Can easily be integrated by seperating,

For $\int\frac{\sqrt{1+x}} {{x}}dx $,

Let,

$1+x = t^2$

$dx=2tdt$

$\therefore I_1=\int\frac{\sqrt{1+x}} {{x}}dx = \int\frac{2t^2}{t^2-1}dt$

After little arranging,

$I_1 = 2[\int dt + \int \frac{dt}{t^2-1}]$

$I_1 = 2[t+\ln \frac{|t-1|}{|t+1|}]+C$

$\frac{I_1}{2}= \sqrt{1+x}+\ln \frac {|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|} + C$

Similarly, $\frac{I_2}{2} = \sqrt{1-x}+\ln \frac {|\sqrt{1-x}-1|}{|\sqrt{1-x}+1|} + C$

$\therefore I = \frac{1}{2}[I_1+I_2]=\sqrt{1+x}+\ln \frac {|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|} + \sqrt{1-x}+\ln \frac {|\sqrt{1-x}-1|}{|\sqrt{1-x}+1|}+C$

$I = \sqrt{1+x} + \sqrt{1-x} + \ln \frac{|(\sqrt{1+x}-1)(\sqrt{1-x}-1)|}{|(\sqrt{1+x}+1)(\sqrt{1-x}+1)|} + C$

Hope this answers your question