Calc1.14

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\int {\frac  {x\,dx}{{\sqrt  {3-2x-x^{2}}}}}

Note \int {\frac  {x\,dx}{{\sqrt  {3-2x-x^{2}}}}}=\int {\frac  {x\,dx}{{\sqrt  {4-(x+1)^{2}}}}}. Now make the trigonometric substitution x+1=2\sin \theta \,, so that dx=2\cos \theta \,d\theta \, and x=2\sin \theta -1\,. Then,

\int {\frac  {x\,dx}{{\sqrt  {4-(x+1)^{2}}}}} =\int {\frac  {(2\sin \theta -1)2\cos \theta \,d\theta }{{\sqrt  {4-4\sin ^{2}\theta }}}}
=\int {\frac  {(2\sin \theta -1)2\cos \theta \,d\theta }{{\sqrt  {4\cos ^{2}\theta }}}}
=\int (2\sin \theta -1)d\theta =-2\cos \theta -\theta +C.

Now, if {\frac  {x+1}{2}}=\sin \theta , it can be shown that \cos \theta ={\frac  {{\sqrt  {3-2x-x^{2}}}}{2}} and \theta =\arcsin \left({\frac  {x+1}{2}}\right), so

\int {\frac  {x\,dx}{{\sqrt  {3-2x-x^{2}}}}}=-\arcsin \left({\frac  {x+1}{2}}\right)-{\sqrt  {3-2x-x^{2}}}+C.