How to evaluate these 3 integrals? $$(1)\,\int \frac{\arctan\sqrt x}{(x +1 )\sqrt x}dx\qquad (2)\,\int \frac{1}{\sin x + 2\cos x + 3}dx\qquad(3)\,\int \frac{1 - \sqrt{x +1 }}{1 + \sqrt[3]{x + 1}}dx$$
I've just solved about 10 tasks successfully, but nothing works for these ones. I suspect there are some methods I don't know about. I'd be very grateful for any help! 
 A: For the first one take the substitution:
$$u=\arctan(\sqrt{x})$$ 
The second one is a bit tricky here you use 
$$u=\tan\left(\frac{x}{2}\right)$$ and the identities 
$$\sin(x)=\frac{2u}{u^2+1} \qquad \cos(x)=\frac{1-u^2}{u^2+1}$$
Tell when you need more hints.
The third one is more ugly, I try to avoid trigonometrics as much as possible. At first take 
$$u=\sqrt[6]{x+1}$$
After this, you should make long division, which gives you a polynomial and a fraction, split up the fraction.
The long division 
$$\frac{u^5-u^8}{u^2+1}= -u^6 +u^4 +u^3 -u^2 +\frac{u-1}{u^2+1}-u+1$$
A: For example:
$$(\arctan\sqrt x)'=\frac{1}{2\sqrt x(1+x)}\Longrightarrow$$
$$\int\frac{\arctan\sqrt x}{\sqrt x(1+x)}dx=2\int\arctan \sqrt x\,d(\arctan\sqrt x)=$$
$$=\frac{\arctan^2\sqrt x}{2}+C$$
A: Since no one's mentioned the third one, I'd recommend trying the substitution
$$x+1=\tan^6u,dx=6\tan^5u\sec^2udu$$
This yields
$$\int6\tan^5u-6\tan^8udu=$$
$$\int(6\tan^3u-6\tan^6u)\sec^2udu-\int6\tan^3u-6\tan^6udu$$
The first half can be evaluated after a simple substitution.  Continue the process, eliminating all powers of tangent over 1.
