If $$\int x^5 e^{-x^2} dx = g(x)e^{-x^2} + c$$, where c is a constant of integration, then g(-1) is equal to

We have been told that the antiderivative can be written in the form

$$\int x^{5}e^{-x^{2}}\,dx = g(x)\,e^{-x^{2}} + c,$$

where $$c$$ is the constant of integration. In order to determine the value of $$g(-1)$$ we will first find an explicit expression for the function $$g(x)$$.

Differentiate both sides with respect to $$x$$. The derivative of the left-hand side is obviously the integrand itself, while on the right-hand side we must apply the product rule:

$$\frac{d}{dx}\!\left[g(x)e^{-x^{2}}\right] = g'(x)\,e^{-x^{2}} \;+\; g(x)\,\frac{d}{dx}\!\bigl(e^{-x^{2}}\bigr).$$

Since $$\dfrac{d}{dx}\bigl(e^{-x^{2}}\bigr)=e^{-x^{2}}\!\cdot(-2x),$$ we obtain

$$g'(x)\,e^{-x^{2}} \;-\; 2x\,g(x)\,e^{-x^{2}} \;=\; x^{5}e^{-x^{2}}.$$

Because the common factor $$e^{-x^{2}}$$ is never zero, we can divide it out, giving the first-order linear differential equation

$$g'(x)\;-\;2x\,g(x)\;=\;x^{5}. \quad -(1)$$

This equation can be solved either by the integrating-factor method or by an intelligent guess that $$g(x)$$ should be a polynomial. A direct polynomial approach is quicker here.

Assume that $$g(x)$$ is a polynomial in $$x$$. The right-hand side of (1) is of degree 5, so let us seek $$g(x)$$ of the form

$$g(x)=A\,x^{4}+B\,x^{2}+C,$$

where $$A,B,C$$ are constants to be determined.

Compute the derivative:

$$g'(x)=4A\,x^{3}+2B\,x.$$

Next, compute the term $$-2x\,g(x)$$ explicitly:

$$-2x\,g(x) = -2x\left(A\,x^{4}+B\,x^{2}+C\right) = -2A\,x^{5}-2B\,x^{3}-2C\,x.$$

Add the two results to get the left side of (1):

$$g'(x)-2x\,g(x) = \bigl(4A\,x^{3}+2B\,x\bigr)\;+\;\bigl(-2A\,x^{5}-2B\,x^{3}-2C\,x\bigr).$$

Collect like powers of $$x$$:

$$$ \begin{aligned} x^{5}&:;;-2A,\\ x^{3}&:;;4A-2B,\\ x^{1}&:;;2B-2C. \end{aligned} $$$

Equation (1) requires this polynomial to equal $$x^{5}$$, i.e. to have coefficient $$1$$ for $$x^{5}$$ and coefficient $$0$$ for every other power. Thus we get the system of equations

$$$ \begin{aligned} -2A &= 1,\\ 4A-2B &= 0,\\ 2B-2C &= 0. \end{aligned} $$$

Solve one by one:

From $$-2A=1$$ we obtain $$A=-\dfrac12.$$

Substituting $$A=-\dfrac12$$ in $$4A-2B=0$$ gives $$4\!\left(-\dfrac12\right)-2B=0 \;\Longrightarrow\; -2-2B=0 \;\Longrightarrow\; B=-1.$$

Using $$B=-1$$ in $$2B-2C=0$$ yields $$2(-1)-2C=0 \;\Longrightarrow\; -2-2C=0 \;\Longrightarrow\; C=-1.$$

Therefore

$$g(x)=A\,x^{4}+B\,x^{2}+C = -\frac12\,x^{4}-x^{2}-1.$$\

Now evaluate this expression at $$x=-1$$:

$$$ \begin{aligned} g(-1) &= -\frac12\,(-1)^{4}-(-1)^{2}-1\\[4pt] &= -\frac12\,(1)-1-1\\[4pt] &= -\frac12-2\\[4pt] &= -\frac52. \end{aligned} $$$

Hence, the correct answer is Option A.