How to find this indefinite integral? Find $\displaystyle\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}$ 
I think I should change dx and let x = something. What is your suggestion?
 A: HINT: Let $x=\tan^2\theta$
$$dx=2\tan\theta \sec^2\theta d\theta$$
$$\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}$$
$$=\int \frac{2\tan\theta \sec^2\theta d\theta}{2\sqrt {\tan^2\theta}+\sqrt{1+\tan^2\theta}+1}$$
$$=\int \frac{2\tan\theta \sec^2\theta d\theta}{2| \tan\theta|+|\sec\theta|+1}$$
you can proceed by assuming $0<\theta< \pi/2$
A: Let me try to use $x=\sinh^2\theta$, then $dx=2\sinh\theta\cosh\theta d\theta=\sinh2\theta d\theta$
$$\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}$$
$$=\int \frac{\sinh2\theta d\theta}{2\sinh\theta+\cosh\theta+1}$$
$$=\int \frac{(e^{2\theta}-e^{-2\theta})/2}{e^\theta+(e^\theta+e^{-\theta})/2+1}d\theta$$
$$=\int \frac{e^{4\theta}-1}{e^\theta(3e^{2\theta}+2e^\theta-1)}d\theta$$
let $y=e^\theta$, $dy=e^\theta d\theta, i.e. d\theta=\frac{1}{y}dy$, the integral becomes
$$\int \frac{y^4-1}{y^2(3y^2+2y-1)}dy$$
$$=\int \frac{(y^2+1)(y-1)}{y^2(3y-1)}dy$$
$$=\int \frac{y^3-y^2+y-1}{y^2(3y-1)}dy$$
$$=\frac13\int [1+\frac{-2y^2+3y-3}{y^2(3y-1)}]dy$$
$$=\frac13\int [1-\frac2{9}\frac{9y^2-2y}{y^2(3y-1)}+\frac1{9}\frac{23y-27}{y^2(3y-1)}]dy$$
$$=\frac y3-\frac2{27}\ln|3y^3-y^2|+\frac1{27}\int[\frac{27}{y^2}-\frac{174}{3y-1}+\frac{58}y]dy$$
$$=\frac y3-\frac2{27}\ln|3y^3-y^2|-\frac1y-\frac{58}{27}\ln|3y-1|+\frac{58}{27}\ln|y|+C$$
$$=\frac y3-\frac1y-\frac{20}{9}\ln|3y-1|+2\ln|y|+C$$
Therefore $\displaystyle\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}=\frac y3-\frac1y-\frac{20}{9}\ln|3y-1|+2\ln|y|+C$, where $y=e^\theta$, $x=\sinh^2\theta$ and $C$ is constant. Also I find that $\theta\in[1+\sqrt2,+\infty)$. Did I write something wrong?
