Need help in indefinite integral elimination

Question

I'm trying to get a easy and short answer for that indefinite integral:

$$\int\frac{\sqrt{x^2+x+1}}{(x+1)^2}\,dx$$

But everything I find is very big and difficult to understand (too much so it seems like a wrong result). Is there any way to solve it in a more beautiful way?

Answer 1

Motivated by completing the square in $x^2+x+1$, the change of variables $u = \sqrt{4/3} (x + 1/2)$ makes the integral $$ \int \dfrac{\sqrt{u^2+1}}{(u + 1/\sqrt{3})^2}\; du $$ Then another substitution $v = \sqrt{u^2+1} - u$ transforms this to the integral of a rational function, which can be done using partial fractions.

EDIT: More generally, the change of variables $v = \sqrt{x^2 + a x + b} - x$ (and thus $x = (v^2-b)/(a-2v)$) is often useful for integrals involving $\sqrt{x^2+ax+b}$. It transforms the integral $$ \int \sqrt{x^2+ax+b}\; R(x)\; dx$$ to $$ \int \dfrac{2(v^2-av+b)^2}{(a-2v)^3} R\left(\frac{v^2-b}{a-2v}\right)\; dv$$

Answer 2

If we let $$p=2\frac{\sqrt{x^2+x+1}}{x+1},q=\frac{x-1}{x+1} $$So $p^2=3+q^2$ with $p\text{d}p=q\text{d}q$.And

\begin{aligned}\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}}&=\frac{1}{2}\int{\frac{q+1}{p}\mathrm{d}q} \\ &=\frac{1}{2}\int{\mathrm{d}p}+\frac{1}{2}\int{\frac{\mathrm{d}q}{p}} \\ &=\frac{1}{2}p+\frac{1}{2}\int{\frac{\mathrm{d}q+\mathrm{d}p}{p+q}} \\ &=\frac{1}{2}p+\frac{1}{2}\ln \left( p+q \right) \end{aligned} At this time,indefinite integral can be easily done: \begin{align}&\int{\frac{\sqrt{x^2+x+1}}{\left( 1+x \right) ^2}\mathrm{d}x}=\int{\frac{x^2+x+1}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}\mathrm{d}x} \\ &=\int{\frac{\mathrm{d}x}{\sqrt{x^2+x+1}}}-\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}} \\ &=\ln \left( x+\frac{1}{2}+\sqrt{x^2+x+1} \right) -\frac{\sqrt{x^2+x+1}}{x+1}-\frac{1}{2}\ln \left( \frac{x-1+2\sqrt{x^2+x+1}}{x+1} \right) \end{align}

Answer 3

Let's calculate by parts first
$\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} + \int{\frac{1}{x+1}\cdot\frac{2x+1}{2\sqrt{x^2+x+1}}dx}\\ \int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} +\int{\frac{(2x+2)-1}{2(x+1)\sqrt{x^2+x+1}}dx}\\ \int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} +\int{\frac{1}{\sqrt{x^2+x+1}}dx} - \frac{1}{2}\int{\frac{1}{(x+1)\sqrt{x^2+x+1}}dx} $

Integral $\int{\frac{1}{\sqrt{x^2+x+1}}dx}$ can be calculated with substitution $\sqrt{x^2+x+1} = u - x$
and integral $\int{\frac{1}{(x+1)\sqrt{x^2+x+1}}dx}$ can be calculated with substitution $\sqrt{x^2+x+1} = u\cdot x - 1$