Solving integral $ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x $

Question

there is integral $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x$$ i am trying to separate this : $$=\int \mathrm{d}x -\int \frac{\mathrm{d}x}{1+x+\sqrt{1+x+x^2}} $$ but have no idea about second

Answer 1

We have the algebraic form: $$ \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}. $$ Multiplying by $\dfrac{(1+x)-\sqrt{1+x+x^2}}{(1+x)-\sqrt{1+x+x^2}}$ yields $$ \frac{\sqrt{1+x+x^2}-1}{x}=\frac{\sqrt{1+x+x^2}}{x}-\frac1x. $$ The integral becomes $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\ dx=\int \frac{\sqrt{1+x+x^2}}{x}\ dx-\int \frac1x\ dx. $$ The left part integral in RHS can be solved using Euler substitution by letting $t-x=\sqrt{1+x+x^2}$, you will get $x=\dfrac{t^2-1}{2t+1}$, $dx=\dfrac{2(t^2+t+1)}{(2t+1)^2}\ dt$, and $\sqrt{x^2+x+1}=\dfrac{t^2+t+1}{2t+1}$, then it becomes $$ \int \frac{\sqrt{1+x+x^2}}{x}\ dx=\int \dfrac{2(t^2+t+1)^2}{(t^2-1)(2t+1)^2}\ dt. $$ The last part can be solved by using partial fraction decomposition $$ \dfrac{2(t^2+t+1)^2}{(t^2-1)(1+2t)^2}=-\frac1{t+1}+\frac1{2t+1}-\frac3{2(2t+1)^2}+\frac1{t-1}+\frac12. $$ I hope I don't mess up and this helps you.

Answer 2

Decompose the integrand as follows \begin{align} &\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}dx\\ =& \int \frac{2x+1}{2\sqrt{x^2+x+1}}+\frac1{2\sqrt{x^2+x+1}}+\frac1x\bigg(\frac1{\sqrt{x^2+x+1}} -1\bigg)\ dx\\ = &\ \sqrt{x^2+x+1}+\frac12\sinh^{-1}\frac{2x+1}{\sqrt3} -\ln \bigg(\frac x2+1+\sqrt{x^2+x+1}\bigg)+C \end{align}