Indefinite integration of $\int \frac{1}{1+\sqrt{x^2+2x+2}}dx$

Question

Integrate $$\int \frac{1}{1+\sqrt{x^2+2x+2}}dx$$

I have tried by using Euler substitution, but that gave me a wrong answer.

So can somebody help?

Answer 1

If you do $x=\tan(y)-1$ and $\mathrm dx=\sec^2(y)\,\mathrm dy$, you get$$\int\frac{\sec^2(y)}{1+\sec(y)}\,\mathrm dy=\int\frac1{\cos(y)+\cos^2(y)}\,\mathrm dy.$$Now, doing $y=2\arctan(\theta)$ and $\mathrm dy=\frac{2\,\mathrm d\theta}{1+\theta^2}$, you get$$\int\frac2{\left(\frac{1-\theta^2}{1+\theta^2}+\left(\frac{1-\theta^2}{1+\theta^2}\right)^2\right)(1+\theta^2)}\,\mathrm d\theta=\int\frac{1+\theta^2}{1-\theta^2}\,\mathrm d\theta.$$Can you take it form here?

Answer 2

With $x+1=\dfrac12\left(u-\dfrac1u\right)$, so that $dx=\dfrac12\left(1+\dfrac1{u^2}\right)du$,

$$I=\int\frac{\dfrac12\left(1+\dfrac1{u^2}\right)}{1+\dfrac12\left(u+\dfrac1u\right)}du=\int\frac{u^2+1}{(u+1)^2}du=u-2\int\frac{u}{(u+1)^2}du,$$ which is easy by parts.

To express $u$ as a function of $x$, you solve a quadratic equation (or use $x+1=\sinh(\log u)$).

Answer 3

Let $\sqrt{x^2+2x+2}=\cosh t$, or $x+1=\sinh t$ \begin{align} \int \frac{1}{1+\sqrt{x^2+2x+2}}dx =& \int \frac{\cosh t}{1+ \cosh t}dt=t-\tanh \frac t2+C \end{align}

Answer 4

A small deviation from user65203's solution to end up with Josè's integral:

$$\begin{align*} \int \frac{dx}{1+\sqrt{x^2+2x+2}} &= \int \frac{dy}{1+\sqrt{y^2+1}} &y=x+1 \\ &= \int \frac{1+z^2}{(1-z^2)^2+z^2(1-z^2)} \, dz & y=\frac{2z}{1-z^2}\\ &= \int \frac{1+z^2}{1-z^2} \, dz \end{align*}$$

(whereas the other solution takes $y=\frac{1-z^2}{2z}$)

Answer 5

In Josè's answer there was a partial fraction, after $x+1=\tan\theta$ substitution, we get $$\small\int \left(\frac{1}{\cos\theta}-\frac{1}{1+\cos\theta}\right)d\theta=\int(\sec\theta-\frac12\sec^2\frac\theta 2)d\theta=\ln|x+1+\sqrt{x^2+2x+2}|-\frac{1-\sqrt{x^2+2x+2}}{x+1}+c.$$