# Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$

$$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$ How to integrate? I tried the substitution $x=\sin\theta$ but didn't work.

• how about $x = \tan \theta?$
– abel
Dec 30, 2014 at 4:17
• if that does not work try $x = \frac{2t}{1+t^2}$
– abel
Dec 30, 2014 at 4:19

I would start by using the substitution $u=\frac{1-x}{1+x}$. This is a handy substitution because it is its own inverse:

$$u=\frac{1-x}{1+x}\implies x=\frac{1-u}{1+u},~\mathrm{d}x=\frac{-2}{(1+u)^2}\,\mathrm{d}u.$$

Then,

\begin{align} \int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x &=\int\left(\frac{1+u}{1-u}\right)\sqrt{u}\frac{(-2)}{(1+u)^2}\,\mathrm{d}u\\ &=-2\int\frac{\sqrt{u}}{(1-u)(1+u)}\,\mathrm{d}u\\ &=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u.\\ \end{align}

Another substitution such as $u=t^2$ would transform this to an integral of a rational function:

\begin{align} \int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x &=-2\int\frac{\sqrt{u}}{1-u^2}\,\mathrm{d}u\\ &=-2\int\frac{t}{1-t^4}\left(2t\,\mathrm{d}t\right)\\ &=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t.\\ \end{align}

From there, partial fraction decomposition will break the integral down into elementary integrals, and back-substitution will yield the desired anti-derivative. PFD gives us,

$$-\frac{4t^2}{1-t^4}=\frac{2}{1+t^2}-\frac{1}{1+t}-\frac{1}{1-t},$$

so,

\begin{align} \int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x &=-4\int\frac{t^2}{1-t^4}\,\mathrm{d}t\\ &=2\int\frac{\mathrm{d}t}{1+t^2}-\int\frac{\mathrm{d}t}{1+t}-\int\frac{\mathrm{d}t}{1-t}\\ &=2\tan^{-1}{\left(t\right)}-\ln{\left(1+t\right)}+\ln{\left(1-t\right)}+\color{grey}{constant}\\ &=2\tan^{-1}{\left(t\right)}-2\tanh^{-1}{\left(t\right)}+\color{grey}{constant}\\ &=2\tan^{-1}{\left(\sqrt{u}\right)}-2\tanh^{-1}{\left(\sqrt{u}\right)}+\color{grey}{constant}\\ &=2\tan^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}-2\tanh^{-1}{\left(\sqrt{\frac{1-x}{1+x}}\right)}+\color{grey}{constant}.~\blacksquare\\ \end{align}

Let $x = \sin \theta$ $$\int \frac{1}{\sin \theta} \cdot \sqrt \frac{1-\sin \theta}{1 + \sin \theta} \cdot \cos \theta \; d \theta \;\;= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \sqrt\frac{(1-\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} \; d\theta \\ = \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \frac{1-\sin\theta}{\cos\theta}\; d\theta \;\; = \;\; \int (\csc\theta \; - 1) \; d\theta$$ Then convert back in terms of x

Observing that $$f(x)=\frac{1}{x}\sqrt{\frac{1-x}{1+x}}=\frac{1}{x}\sqrt{\frac{1-x}{1+x}\cdot\frac{1-x}{1-x}}=\frac{1}{x}\frac{1-x}{\sqrt{1-x^2}}$$ so we have $$I=\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\mathrm d x=\underbrace{\int \frac{1}{x\sqrt{1-x^2}}\mathrm d x}_J-\underbrace{\int \frac{1}{\sqrt{1-x^2}}\mathrm d x}_K=J-K$$ Putting $x=\sin u$ we have $$K=\int \mathrm d u=u+C_1$$ that is $K=\arcsin x+ C_1$; and for $J$ $$J=\int \frac{1}{\sin u}\mathrm d u$$ Putting $t=\tan(\frac{u}{2})$, so that $\sin u=\frac{2t}{1+t^2}$ and $\mathrm{d}u=\frac{2}{1+t^2}\mathrm{d}t$ we find $$J=\int \frac{1}{t}\mathrm{d}t=\log t+C_2$$ Substituting back $t=\tan(\frac{u}{2})=\tan\left(\frac{\arcsin(x)}{2}\right)$ we have $$J=\log \left(\tan\left(\frac{\arcsin(x)}{2}\right)\right)+C_2$$ Finally $$I=\arcsin x+\log \left(\tan\left(\frac{\arcsin(x)}{2}\right)\right)+C$$

You could also use $$\sqrt\frac{1-x}{1+x}=y$$ that is to say $$x=\frac{1-y^2}{1+y^2}$$ $$dx=-\frac{4 y}{\left(1+y^2\right)^2}\space dy$$ So,$$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx=\int\frac{4 y^2}{y^4-1}\space dy$$ Now, partial fraction decomposition leads to $$\frac{4 y^2}{y^4-1}=\frac{2}{y^2+1}-\frac{1}{y+1}+\frac{1}{y-1}$$

I am sure that you can take from here.

Please, notice that this is very similar to what David H. proposed as a solution.

i will try $x = {2t \over 1 + t^2}$
$$dx = 2{ (1+t^2)dt - t \cdot2tdt \over (1+t^2)^2} = {2(1-t^2)dt \over (1+t^2)^2} \\ 1 - x = {(1-t)^2 \over 1 + t^2}, 1 + x = {(1+t)^2 \over 1 + t^2}, \sqrt{{1-x \over 1 + x}}= {1 - t \over 1 + t}$$

putting all these together,

$$\int{1 \over x} \sqrt{{1-x \over 1 + x}} dx = \int{1 + t^2 \over 2t} {1 - t \over 1 + t} {2(1-t^2)dt \over (1+t^2)^2} \\= \int {(1-t)^2 \over t(1+t^2)} \ dt = \int {1 + t^2 - 2t \over t(1+t^2)} dt = \int {dt \over t} - 2\int{dt \over 1 + t^2} \\ = \ln(t) - 2 \tan^{-1}(t) + C$$

you can solve $$x = {2t \over 1 + t^2}$$ for $x$ by writing $$xt^2 - 2t + x =0$$ as a quadratic equation for $t$. by quadratic formula we get $$t = {1 \pm \sqrt{1 + x^2} \over x}$$

put all these in $$\int{1 \over x} \sqrt{{1-x \over 1 + x}} dx = \ln(t) - 2 \tan^{-1}(t) + C$$

Solution

$$\text{substitute x with tan }\theta \text{ as pointed out by abel}$$

Now

$$\int\frac{1}{\tan \theta}\sqrt\frac{1 -\tan\theta}{1+\tan \theta}\ dx = \int \cot \sqrt{\frac{(1-\tan \theta)^2}{(1+\tan\theta) (1 -\tan\theta)}}\sec^2\theta$$

Then

$$\int \cot\sec^2 \frac{1-\tan\theta}{\sec\theta} = \int{\cot\sec (1-\tan\theta}) = \int\frac{\sec (1-\tan\theta)}{\tan\theta} = \int\frac{\sec\theta}{\tan\theta} -1$$

Now you can find integral of this.

• $1-\tan^2 \theta \neq \sec^2 \theta$ Dec 30, 2014 at 5:03
• Rather sloppy writing.
– MasB
Dec 30, 2014 at 5:07