Question

$\int e^x(\frac{x^4+x^3-3x^2+x-2}{(x-1)^2})dx=e^x(\frac{f(x)}{x-1})+C$ then $f(2)$ is equal to ($f(x)$ is a polynomial)

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

Answer: (C)

9

Answer 2

The integral is of the form $\int e^x G(x) dx = e^x H(x) + C$, where $G(x) = \frac{x^4+x^3-3x^2+x-2}{(x-1)^2}$ and $H(x) = \frac{f(x)}{x-1}$. This implies $G(x) = H(x) + H'(x)$. We can decompose $G(x)$ using polynomial long division or substitution. Let $u = x-1$, so $x = u+1$. $G(x) = \frac{(u+1)^4+(u+1)^3-3(u+1)^2+(u+1)-2}{u^2}$ $G(x) = \frac{u^4+5u^3+6u^2+2u-2}{u^2} = u^2+5u+6 + \frac{2u-2}{u^2}$ Substituting back $u=x-1$: $G(x) = (x-1)^2+5(x-1)+6 + \frac{2(x-1)-2}{(x-1)^2}$ $G(x) = (x^2-2x+1)+(5x-5)+6 + \frac{2x-4}{(x-1)^2}$ $G(x) = x^2+3x+2 + \frac{2x-4}{(x-1)^2}$ We need to find $H(x)$ such that $H(x)+H'(x) = G(x)$. Let $H(x) = P(x) + \frac{K}{x-1}$, where $P(x)$ is a polynomial. Then $H'(x) = P'(x) - \frac{K}{(x-1)^2}$. $H(x)+H'(x) = P(x)+P'(x) + \frac{K}{x-1} - \frac{K}{(x-1)^2}$. Comparing with $G(x) = x^2+3x+2 + \frac{2x-4}{(x-1)^2}$: $P(x)+P'(x) = x^2+3x+2$. If $P(x)=x^2+x+1$, then $P'(x)=2x+1$, and $P(x)+P'(x) = x^2+3x+2$. $\frac{K}{x-1} - \frac{K}{(x-1)^2} = \frac{2x-4}{(x-1)^2}$. Multiplying by $(x-1)^2$, we get $K(x-1)-K = 2x-4$, which simplifies to $Kx-2K = 2x-4$. Thus, $K=2$. So, $H(x) = x^2+x+1 + \frac{2}{x-1}$. Given $H(x) = \frac{f(x)}{x-1}$: $\frac{f(x)}{x-1} = x^2+x+1 + \frac{2}{x-1} = \frac{(x^2+x+1)(x-1)+2}{x-1}$ $f(x) = (x^2+x+1)(x-1)+2 = (x^3-1)+2 = x^3+1$. We need to find $f(2)$: $f(2) = (2)^3+1 = 8+1 = 9$.