Question

$- {loglog}x + c$

• A. $\int_{}^{}{\frac{1}{\sqrt{x}}\tan^{4}\sqrt{x}}\sec^{2}\sqrt{x}\mspace{6mu} dx =$
• B. $2\tan^{5}\sqrt{x} + c$
• C. $\frac{1}{5}\tan^{5}\sqrt{x} + c$
• D. $\frac{2}{5}\tan^{5}\sqrt{x} + c$

Answer 1

Answer: (C)

$\frac{1}{5}\tan^{5}\sqrt{x} + c$

Answer 2

Put $\int_{}^{}{\frac{x - 1}{(x + 1)^{2}}\mspace{6mu} dx =}$, therefore

$\log(x + 1) + \frac{2}{x + 1} + c$

$\log(x + 1) - \frac{2}{x + 1} + c$.