<s>[INST] Find $\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$ I came across the integral $$\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$$ in a calculus textbook. At this point in the book, only u-substitutions were covered, which brings me to think that there is a clever substitution that one can use to knock off this integral.
I was able to find the answer using $x= 2 \sin \theta$, doing a bit of polynomial long division and then Weiestrass substitution. However, this solution was rather ugly and I don't think this was what the author had in mind.
What else could I try here? Wolfram gives a closed form $$\pi + \sqrt{5} \left [ \tan ^{-1} \left ( \frac{1}{\sqrt{15}} \right ) - \tan ^{-1} \left ( \frac{7}{\sqrt{15}} \right ) \right ]$$
and the indefinite integral was 
$$\sqrt{4-x^2} - \sqrt{5} \tan ^{-1} \left ( \frac{3x+4}{\sqrt{5}\sqrt{4-x^2}}\right ) + 3 \sin ^{-1} \left ( \frac{x}{2} \right )+C $$
 [/INST] HINT:
Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$$I+I=6\int_{-1}^1\dfrac{\sqrt{4-x^2}}{9-x^2}dx$$
Putting $x=2\sin y\implies$
$$2I=6\int_{-\pi/6}^{\pi/6}\dfrac{4\cos^2y}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\dfrac{5+4(1-\sin^2y)-5}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\left(1-\dfrac5{9-\sin^2y}\right)dy$$
Now, $$\int\dfrac{dy}{9-\sin^2y}=\int\dfrac{\csc^2y}{9(1+\cot^2y)-1}dy$$
Set $\cot y=u$
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