1.

What is ∫ tan-1 (sec x + tan x) dx equal to?

A. \(\frac{{\pi x}}{4} + \frac{{{x^2}}}{4} + c\)
B. \(\frac{{\pi x}}{2} + \frac{{{x^2}}}{4} + c\)
C. \(\frac{{\pi x}}{4} + \frac{{\pi {x^2}}}{4} + c\)
D. \(\frac{{\pi x}}{4} - \frac{{{x^2}}}{4} + c\)
Answer» B. \(\frac{{\pi x}}{2} + \frac{{{x^2}}}{4} + c\)


Discussion

No Comment Found