Question

$\int_{}^{}\frac{\mathbf{dx}}{\mathbf{(x + 1)}\sqrt{\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{1}}}$ is equal to

• A. $\sqrt{\frac{\mathbf{x + 1}}{\mathbf{x - 1}}}$+ c
• B. $\mathbf{2}\sqrt{\frac{\mathbf{x}\mathbf{-}\mathbf{1}}{\mathbf{x + 1}}}$ + c
• C. $\sqrt{\frac{\mathbf{x - 1}}{\mathbf{x + 1}}}$+ c
• D. $\mathbf{2}\sqrt{\frac{\mathbf{x + 1}}{\mathbf{x}\mathbf{-}\mathbf{1}}}$+ c

Answer 1

Answer: (C)

$\sqrt{\frac{\mathbf{x - 1}}{\mathbf{x + 1}}}$+ c

Answer 2

$\int \frac { d x } { ( x + 1 ) ^ { 3 / 2 } ( x - 1 ) ^ { 1 / 2 } } = \int \frac { d x } { \left( \frac { x - 1 } { x + 1 } \right) ^ { 1 / 2 } ( x + 1 ) ^ { 2 } }$ put $\frac{x - 1}{x + 1}$= t