Is there easier way to take $\int^3_0 \mathrm{arcsin} \sqrt{\frac{x}{1+x}} dx$? Actually, the integral
$$\int^3_0 \mathrm{arcsin} \sqrt{\frac{x}{1+x}} dx$$
Should be taken by parts, but isn't there an alternative way to solve it ? Otherwise calculus seem to be quite long.
P.S. the problem itself is from Demidovich, so illusions of complicated problems are possible, that's why I am asking.
 A: The integrand function is increasing from $0$ to $\frac{\pi}{3}$ on the integration range and its inverse function is given by $\tan^2(y)$. In particular
$$ \int_{0}^{3}\arcsin\sqrt{\frac{x}{1+x}} = \pi-\int_{0}^{\pi/3}\tan^2(y)\,dy =\pi-\int_{0}^{\sqrt{3}}\frac{t^2}{1+t^2}\,dt$$
and the answer is clearly given by $\color{red}{\frac{4\pi}{3}-\sqrt{3}}$.
A: This is a way to go. First perform the substitution $y=\sqrt{x}$. You obtain
$$\int_0^{\sqrt{3}}\!dy\,2y\arcsin\frac{y}{\sqrt{1+y^2}}
=2\int_0^{\sqrt{3}}\!dy\,y\arctan y.$$
Now you can integrate by parts to obtain the result:
$$2\int_0^{\sqrt{3}}\!dy\,y\arctan y =  \underbrace{y^2 \arctan(y)\Big|_{y=0}^{\sqrt{3}}}_{=\pi} - \int_0^\sqrt{3}\!dy\, \underbrace{\frac{y^2}{1+y^2}}_{1 -\frac{1}{1+y^2}} = \pi+ (\arctan y- y)\Big|_{y=0}^{\sqrt{3}}  = \frac{4\pi}{3}-\sqrt{3}.$$
A: Let $x=\tan^2{\theta}$. (Why? because of the simplification this gives to the thing inside the arcsine, as we shall see...) Then $dx = 2\tan{\theta}\sec^2{\theta} \, d\theta$, while
$$ \sqrt{\frac{x}{1+x}} = \sqrt{\frac{\tan^2{\theta}}{1+\tan^2{\theta}}} = \sqrt{\frac{\tan^2{\theta}}{\sec^2{\theta}}} = \sqrt{\sin^2{\theta}}. $$
On the interval considered, $0<x<3$, $0<x<\arctan{\sqrt{3}} = \pi/3$, so this is just $\sin{x}$ since $\sin{x}>0$ on this interval. So the integral becomes
$$ \int_{0}^{\pi/3} \arcsin{\sin{\theta}} (\tan^2{\theta})' \, d\theta = \int_{0}^{\pi/3} \theta (\tan^2{\theta})' \, d\theta $$
since $\arcsin{\sin{\theta}} = \sin{\theta}$ for $-\pi/2<\theta<\pi/2$. Integration by parts gives
$$ \frac{\pi}{3}\tan^2{(\pi/3)} - \int_0^{\pi/3} \tan^2{\theta} \, d\theta, $$
which one can finish off using $\tan^2{\theta}=\sec^2{\theta}-1$.
A: What about making $t=\arcsin\sqrt{\frac x{x+1}}$, then $$\sin^2 t=\frac x{x+1}\quad\implies\quad x(1-\sin^2t)=\sin^2 t\quad\implies\quad x=\frac{\sin^2 t}{1-\sin^2 t}=\tan^2 t$$
So $dx=2\tan t\sec^2t\,dt$, and
\begin{align*}
\int_0^3\arcsin\sqrt{\frac x{x+1}}\,dx&=\int_0^{\pi/3}2t\tan t\sec^2 t\,dt\\[3pt]
&=\left.t\tan^2 t\right|_0^{\pi/3}-\int_0^{\pi/3}\tan^2 tdt
\end{align*}
