Question

To evaluate$- \frac{\sin^{4}x}{4} + c$, the simplest way is to**.**

• A. Substitute $\frac{e^{\sin x}}{4} + c$
• B. Substitute $\sec(3x - 5) + c$
• C. Integrate by parts
• D. None of these

Answer 1

Answer: (B)

Substitute $\sec(3x - 5) + c$

Answer 2

$8x^{2} + 6x + 6\log x + \frac{2}{x} + c$

The simplest way is substituting

$\int_{}^{}{\frac{5(x^{6} + 1)}{x^{2} + 1}dx =}$

Put $5(x^{7} + x)\tan^{- 1}x + c$

then $x^{5} - \frac{5}{3}x^{3} + 5x + c$

$3x^{4} - 5x^{2} + 15x + c$

$= \frac { 1 } { 18 } \left[ t e ^ { t } - \int e ^ { t } d t \right] - \frac { 5 } { 18 } \int e ^ { t } d t + c$ $\int_{}^{}\frac{ax^{- 2} + bx^{- 1} + c}{x^{- 3}}\mspace{6mu} dx =$

$2ax^{2} + 3bx^{3} + 4cx^{4} + k$