Question

$\int_{}^{}\frac{\mathbf{x + 2}}{\sqrt{\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{2x + 4}}}$equals

• A. $\sqrt{x^{2} - 2x + 4} + 3{\sin h}^{- 1}\left\lbrack \frac{(x - 1)}{\sqrt{3}} \right\rbrack + c$
• B. $\sqrt{x^{2} - 2x + 4} - 3{\sin h}^{- 1}\left\lbrack \frac{(x - 1)}{\sqrt{3}} \right\rbrack + c$
• C. $\sqrt{x^{2} - 2x + 4} + 3{\cos h}^{- 1}\left\lbrack \frac{(x - 1)}{\sqrt{3}} \right\rbrack + c$
• D. $\sqrt{x^{2} - 2x + 4} - 3{\cos h}^{- 1}\left\lbrack \frac{(x - 1)}{\sqrt{3}} \right\rbrack + c$

Answer 1

Answer: (A)

$\sqrt{x^{2} - 2x + 4} + 3{\sin h}^{- 1}\left\lbrack \frac{(x - 1)}{\sqrt{3}} \right\rbrack + c$

Answer 2

$\int_{}^{}\frac{\mathbf{x + 2}}{\sqrt{\mathbf{x}^{\mathbf{2}}\mathbf{- 2x + 4}}}\mathbf{dx =}\int_{}^{}{\frac{\mathbf{x + 2}}{\sqrt{\mathbf{(x - 1}\mathbf{)}^{\mathbf{2}}\mathbf{+}\left( \sqrt{\mathbf{3}} \right)^{\mathbf{2}}}}\mathbf{dx =}\int_{}^{}{\frac{\mathbf{x - 1 + 3}}{\sqrt{\mathbf{(x - 1}\mathbf{)}^{\mathbf{2}}\mathbf{+}\left( \sqrt{\mathbf{3}} \right)^{\mathbf{2}}}}\mathbf{dx}}}\mathbf{=}\int_{}^{}\frac{\mathbf{x}\mathbf{-}\mathbf{1}}{\sqrt{\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{2x + 4}}}\mathbf{dx +}\int_{}^{}\frac{\mathbf{3dx}}{\sqrt{\mathbf{(x}\mathbf{-}\mathbf{1}\mathbf{)}^{\mathbf{2}}\mathbf{+ (}\sqrt{\mathbf{3}}\mathbf{)}^{\mathbf{2}}}}$

Put $x^{2} - 2x + 4 = t^{2}$ in the first expression

$\Rightarrow 2(x - 1)dx = 2tdt$

$\Rightarrow (x - 1)dx = tdt$

$\int_{}^{}\frac{x + 2}{\sqrt{x^{2} - 2x} + 4} ⥂ ⥂ dx$ $= \int_{}^{}{\frac{tdt}{t} + 3\int_{}^{}\frac{dx}{\sqrt{(x - 1)^{2} + (\sqrt{3})^{2}}}}$

$\int_{}^{}{\frac{x + 2}{\sqrt{x^{2} - 2x + 4}}dx} = \sqrt{x^{2} - 2x + 4} + 3{\sin h}^{- 1}\left\lbrack \frac{x - 1}{\sqrt{3}} \right\rbrack + c$