Integration by Parts — Question 8

Question 8 – Solution

We want to choose $P(x)$ so that differentiating $e^xP(x)$ produces the given expression. By the product rule, $$\begin{align*} \frac{d}{dx}\left[e^xP(x)\right] &=e^xP(x)+e^xP'(x) \\ &=e^x\bigl(P(x)+P'(x)\bigr). \end{align*}$$ Thus, the required identity $$\frac{d}{dx}\left[e^xP(x)\right]=e^x(x^2+1)$$ will hold if $$P(x)+P'(x)=x^2+1. \tag{1}$$

Because $P$ is quadratic, write $$P(x)=ax^2+bx+c,$$ where $a$, $b$, and $c$ are constants. Then $$P'(x)=2ax+b.$$ Adding $P(x)$ and $P'(x)$ gives $$\begin{align*} P(x)+P'(x) &=(ax^2+bx+c)+(2ax+b) \\ &=ax^2+(b+2a)x+(c+b). \end{align*}$$ Equation (1) requires this polynomial to equal $$x^2+1=1x^2+0x+1.$$ Matching coefficients of like powers of $x$ yields $$\begin{array}{rcll} a&=&1, & \text{coefficient of }x^2,\\ b+2a&=&0, & \text{coefficient of }x,\\ c+b&=&1, & \text{constant term}. \end{array}$$ Solve the equations in order. Since $a=1$, $$b+2(1)=0 \qquad\Longrightarrow\qquad b=-2.$$ Then $$c+(-2)=1 \qquad\Longrightarrow\qquad c=3.$$ Therefore, $$\boxed{P(x)=x^2-2x+3}.$$

Finally, verify the result. Since $P'(x)=2x-2$, $$\begin{align*} \frac{d}{dx}\left[e^x(x^2-2x+3)\right] &=e^x(x^2-2x+3)+e^x(2x-2) \\ &=e^x\bigl(x^2-2x+3+2x-2\bigr) \\ &=e^x(x^2+1), \end{align*}$$ which is the required expression.