To solve the integral of $-\frac{x}{e^{x^2}} \, dx$, we can apply a substitution method.

Step 1: Substitution

Let: $u = x^2$ Then, differentiate both sides with respect to $x$: $du = 2x \, dx$ Thus, $x \, dx = \frac{1}{2} \, du$.

Step 2: Rewrite the Integral

Now substitute $u = x^2$ and $x \, dx = \frac{1}{2} \, du$ into the original integral: $\int -\frac{x}{e^{x^2}} \, dx = -\frac{1}{2} \int \frac{1}{e^u} \, du$

Step 3: Simplify the Integral

The integral $\int \frac{1}{e^u} \, du$ is simply $\int e^{-u} \, du$, which is: $\int e^{-u} \, du = -e^{-u}$

Step 4: Back-substitute $u = x^2$

Now, substitute $u = x^2$ back into the result: $-\frac{1}{2} \int e^{-u} \, du = \frac{1}{2} e^{-x^2} + C$

Final Answer:

$\int -\frac{x}{e^{x^2}} \, dx = \frac{1}{2} e^{-x^2} + C$

Would you like a more detailed breakdown or have any other questions?

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Here are 5 related questions to explore:

1. What is the integral of $\frac{1}{e^{x^2}}$?
2. How does substitution help simplify integrals involving exponential functions?
3. Can integration by parts be used for this problem?
4. How do you evaluate improper integrals involving exponential decay?
5. What is the difference between indefinite and definite integrals?

Tip: Always check if substitution can simplify an integral before using more advanced techniques!