Question

$\int \frac{1}{\sqrt{8+2x-x^2}}dx=$

• A. $\frac{1}{3}{\sin }^{-1}\left(\frac{x-1}{3}\right)+c$
• B. ${\sin }^{-1}\left(\frac{x+1}{3}\right)+c$
• C. $\frac{1}{3}{\sin }^{-1}\left(\frac{x+1}{3}\right)+c$
• D. ${\sin }^{-1}\left(\frac{x-1}{3}\right)+c$

Answer 1

The correct option is D ${\sin }^{-1}\left(\frac{x-1}{3}\right)+c$

$\int \frac{1}{8+2x-x^2}dx⟹\frac{1}{3}\int \frac{1}{\left(\frac{8+2x-x^2}{3}\right)}dx⟹\frac{1}{3}\int \frac{1}{(1-({\frac{x-1}{3}}^{}{)}^2)}dx⟹\frac{1}{3}{sin}^{-1}\left(\frac{x-1}{3}\right)\left(3\right)+c⟹{sin}^{-1}\left(\frac{x-1}{3}\right)+c$