Question

$\int e^x\left(\frac{x^4+x^2+1}{x^2+x+1}\right)dx=$

• A. $e^x(x^4+x^2+1)+c$
• B. $e^x(x^2+x+1)+c$
• C. $e^x(x^2-3x+4)+c$
• D. $e^x(x^2-4x+5)+c$

Answer 1

Answer: (C)

$e^x(x^2-3x+4)+c$

$\int e^x\left(\frac{x^4+x^2+1}{x^2+x+1}\right)dx$
$\int e^x(x^2-x+1)dx$
$x^2-x+1=(a{x}^2+bx+c)+(2ax+b)$
$a=1;b+2=-1b=-3,c+b=1c=4$
$\int e^x(x^2-x+1)dx=\int e^x\left[\right(x^2-3x+4)+(2x-3\left)\right]$
In the form of $\int e^x\left(f\right(x)+f^1(x\left)\right]dx=e^{x}f\left(x\right)+c$
$=e^x(x^2-3x+4)+c$