2 integrals using Euler substitution

Question

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

I tried with the Euler substitution.I wrote that $\sqrt{x^{2}+x+1}=x+t$ and I got $x=\frac{t^{2}-1}{1-2t}$.I replaced this x in the initial integral and also I replaced dx with x' but I get an ugly result.

The same I get for $$\int \frac{1}{1+\sqrt{x^{2}+2x+2}}dx$$

For the second integral I get $x=\frac{t^{2}-2}{2-2t}$ but when I replace x in the integral I get some ugly result.

How to approach these integrals?

Answer 1

Using Euler substitution as you did, I suppose that you did not simplify enougth since $$ \frac{dx}{(1+x)\sqrt{x^{2}+x+1}}=-2\frac{ \text{sgn}(1-2 t)}{(t-2) t}\,dt$$

Answer 2

A simpler method for $I_1$ is to take $$x+1=1/t \implies dx= -t^{-2}dt \implies I_1= \int \frac{-dt}{\sqrt{t^2-t+1}}=\sinh ^{-1} \frac{3t-1}{\sqrt{3}}= \sinh^{-1} \frac{2-x}{\sqrt{3}(x+1)}+C$$

Answer 3

For the first integral make the substitutions $$x+\frac{1}{2}=\frac{\sqrt 3}{2}\tan t$$and then $$u=\tan \ (\frac {t}{2})$$. For the second integral make the substitution $x+1=\tan t$