Question

$- 2\sqrt{1 - x} + c$

• A. $- \sin^{- 1}\sqrt{x} + c$
• B. $\sin^{- 1}\sqrt{x} + c$
• C. $\int_{}^{}{\frac{dx}{1 - x^{2}} =}$
• D. $\tan^{- 1}x + c$

Answer 1

Answer: (B)

$\sin^{- 1}\sqrt{x} + c$

Answer 2

$\int_{}^{}{\frac{\sec^{2}x\mspace{6mu} dx}{\sqrt{\tan^{2}x + 4}} =}$

Put $\log\left\lbrack \tan x + \sqrt{\tan^{2}x + 4} \right\rbrack + c$ and $\frac{1}{2}\log\left\lbrack \tan x + \sqrt{\tan^{2}x + 4} \right\rbrack + c$ then it

reduces to $\log\left\lbrack \frac{1}{2}\tan x + \frac{1}{2}\sqrt{\tan^{2}x + 4} \right\rbrack + c$

$\int_{}^{}\frac{2x\tan^{- 1}x^{2}}{1 + x^{4}}\mspace{6mu} dx =$