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

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.

• For direct questions on integration, try this - integral-calculator.com Commented Nov 20, 2023 at 10:54

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?

• @Syamaprasad Chakrabarti the edit is wrong Commented Nov 18, 2023 at 16:19

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.

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}

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?

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