Question

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

• A. x tan x – log |sec x| $\mathbf{-}\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+ c}$
• B. x tan x – log |sec x| +$\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+ c}$
• C. x tan x + log |sec x| $\mathbf{-}\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+ c}$
• D. x tan x + log |sec x| + $\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+ c}$

Answer 1

Answer: (A)

x tan x – log |sec x| $\mathbf{-}\frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+ c}$

Answer 2

I =dx = $\int_{}^{}x$tan²x dx

use integration by parts