Question

$I$ : $\int e^x(1-\cot x+{\cot }^{2}x)dx=-e^x\cot x+c$

$II:\int e^x\left(\frac{1+x\log x}{x}\right)dx=e^x\log x+c$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II are true

Answer 1

Answer: (C)

both I and II are true

$SI=\int e^x(cose{c}^{2}x-cotx)dx$
$=-\int e^x(cotx-cose{c}^{2}x)dx$
$=-e^{x}cotx+c$
$SII=\int e^x(\frac{1}{x}+logx)dx=e^{x}logx+c$
Both are in the form of
$\int e^x\left(f\right(x)+f^1(x\left)\right]dx=e^{x}f\left(x\right)+c$
both are true