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

Question

How to integrate $\int \frac{1}{1+{\sqrt{\frac{1}{1+x}}}}dx$? I am trying this integral by substituting $x=\tan^{2}\theta$. Therefore $dx=2(\tan\theta)(\sec^{2}\theta)d\theta$.

Now $\sqrt{\frac{1}{1+x}}=|\cos\theta|$. Now if I take $|\cos\theta|=\cos\theta$, then the result of the above integral will be $\int{(\sec^{2}\frac{\theta}{2})}(\tan\theta)(\sec^{2}\theta)d\theta$.

And for $|\cos\theta|=-\cos\theta$, the result of the above integral will be $\int (\csc^{2}\frac{\theta}{2})(\tan\theta)(\sec^{2}\theta)d\theta$. Thus two different integrals are occurring.

Therefore I thought of taking $|\cos\theta|=(\cos\theta)\text{sgn}(\cos\theta)$. Therefore the above integral will be $\int \frac{1}{1+(\cos\theta)(\text{sgn}(\cos\theta))}d\theta$.

But I can't simplify further. Also I can't understand the case when $|\cos\theta|=0$. Because, when $\cos\theta=0$, then $\tan\theta=\infty$. Therefore, I am in very much doubt with this integral. Please help me out with this integral.

Answer 1

The integral is equivalent to: $$\int \frac{\sqrt{1+x}}{1+\sqrt{1+x}}\mathrm dx=\int \frac{\sqrt u}{1+\sqrt u}\mathrm du $$ Now let $1+\sqrt u=t$, $\mathrm du=2\sqrt u\,\mathrm dt$ and the integral becomes: $$\int \frac{\sqrt u}{1+\sqrt u}\mathrm du=2\int\frac{(t-1)^2}{t}\mathrm dt $$ Can you take it from here?

Answer 2

Let $u=x+1$,

The integrand will become $$\int \frac{1}{\sqrt{\frac{1}{u}}+1}du$$

Let $v=\sqrt{u}$, it then will become $$2\int\frac{v^2}{v+1}dv$$

Do the long division then. I bet you can proceed from here.

Answer 3

Rewriting the integral as $$ I=\int \frac{1}{1+\sqrt{\frac{1}{1+x}}} d x=\int \frac{\sqrt{1+x}}{1+\sqrt{1+x}} d x=x-\int \frac{1}{1+\sqrt{1+x}} d x $$ Putting $y=1+\sqrt{1+x}\Leftrightarrow x=y^2-2y $ transforms the integral into $$ \begin{aligned} I & =x-\int \frac{1}{y}(2 y-2) d y \\ & =x-2 y+2 \ln y \\ & =x-2(1+\sqrt{1+x})+2 \ln (1+\sqrt{1+x})+c \\ & =x-2 \sqrt{1+x}+2 \ln (1+\sqrt{1+x})+C \end{aligned} $$

Answer 4

HINT

Additionally to the other answers, you can also apply hyperbolic substitution: \begin{align*} \int\frac{\sqrt{x + 1}}{1 + \sqrt{x + 1}}\mathrm{d}x & = \int\frac{\sqrt{\sinh^{2}(u) + 1}}{1 + \sqrt{\sinh^{2}(u) + 1}}\mathrm{d}(\sinh(u))\\\\ & = \int\frac{\cosh^{2}(u)}{1 + \cosh(u)}\mathrm{d}u\\\\ & = \int(\cosh(u) - 1)\mathrm{d}u + \int\frac{\mathrm{d}u}{1 + \cosh(u)}\\\\ & = \sinh(u) - u + \int\frac{2}{(e^{u/2} + e^{-u/2})^{2}}\mathrm{d}u \end{align*}

where the last integral can be computed as follows: \begin{align*} \int\frac{\mathrm{d}u}{(e^{u/2} + e^{-u/2})^{2}} & = \int\frac{e^{u}}{(e^{u} + 1)^{2}}\mathrm{d}u\\\\ & = \int\frac{\mathrm{d}(e^{u} + 1)}{(e^{u} + 1)^{2}}\\\\ & = -\frac{1}{e^{u} + 1} + C \end{align*}

Can you take it from here?

Answer 5

$$I=\int \frac{dx}{1+{\sqrt{\frac{1}{1+x}}}}$$ Get rid of the radical $$\sqrt{\frac{1}{1+x}}=t \implies x=\frac{1-t^2}{t^2}\implies dx=-\frac{2}{t^3}\,dt$$ $$I=-2\int \frac{dt}{t^3 (t+1)}=-2\int \left(\frac{1}{t^3}-\frac{1}{t^2}-\frac{1}{t+1}+\frac{1}{t} \right)\,dt $$