allery$\int \dfrac{dx}{\sqrt{2x-x^2}}$ equals
A$\sin^{-1}(x-1)+C$
B$\log\left|x-1+\sqrt{2x-x^2}\right|+C$
C$\tan^{-1}(x-1)+C$
D$\sin^{-1}(x+1)+C$
Answer & Solution
Correct answer: A. $\sin^{-1}(x-1)+C$
1. Complete the square: $2x-x^2=1-(x-1)^2$.
2. Put $t=x-1$, $dt=dx$.
3. Integral $=\int \dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$.
4. Back-substitute: $\sin^{-1}(x-1)+C$. Option D uses the wrong shift.
_Source: syllabus Mathematics Ch 7 "Integrals", p.310_
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