<s>[INST] $ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
 &= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
 &= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\
 &= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{2+2 x+4 \sqrt{x+1}} \cdot d x\\&=\int \frac{1}{1+x+2 \sqrt{x+1}} \cdot d x\\
 &\quad +\int \frac{\sqrt{x+1}}{2(1+x)+4 \sqrt{x+1}}\cdot d x \\
 &\quad -\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot d x
\end{split}
$$
From this last line i can evaluate the first two antiderivatives but the last one $$\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot dx$$ seems a bit hard for me two evaluate, i don't find the good substitution. I am opened to any suggestion. Thanks in advance!
 [/INST] hint
$$x+1=(x-1)+2$$
$$3-x=2-(x-1)$$
put
$$x-1=2\cos(2t)$$
to get
$$x+1=4\cos^2(t)$$
and
$$3-x=4\sin^2(t)$$
the integrale becomes
$$\int\frac{-2\sin(2t)dt}{1+\sin(t)+\cos(t)}$$
which can be computed using the substitution
$$u=\tan(\frac t2)$$
</s>