Question

If $2x^{2} + 8x + 3\log x - \frac{2}{x} + c$ and $8x^{2} + 6x + 6\log x + \frac{2}{x} + c$, then $\int_{}^{}{\frac{5(x^{6} + 1)}{x^{2} + 1}dx =}$

• A. $5\tan^{- 1}(x^{2} + 1) + \log(x^{2} + 1) + c$
• B. $\int_{}^{}\frac{ax^{- 2} + bx^{- 1} + c}{x^{- 3}}\mspace{6mu} dx =$
• C. $2ax^{2} + 3bx^{3} + 4cx^{4} + k$
• D. $6ax^{2} + 4bx^{3} + 3cx^{4} + k$

Answer 1

Answer: (A)

$5\tan^{- 1}(x^{2} + 1) + \log(x^{2} + 1) + c$

Answer 2

$\log 3(3)^{x^{3}} + c$

Put $\int_{}^{}{\sec^{2/3}x\text{cose}\text{c}^{4/3}x\mspace{6mu} dx =}$ then $- 3(\tan x)^{1/3} + c$

Therefore, $- 3(\tan x)^{- 1/3} + c$