Let’s analyze and solve the integrals from the image.

3. $\int_{-\infty}^0 \frac{x}{(x^2 + 2)^4} \, dx$

Step 1: Analyze the integrand

The function $f(x) = \frac{x}{(x^2 + 2)^4}$:

• The numerator is odd ($x$), and the denominator is even ($(x^2 + 2)^4$).
• This makes $f(x)$ odd: $f(-x) = -f(x)$.

Since the integral is over a symmetric interval $[-\infty, 0]$ and the integrand is odd, the result is: $\boxed{0 \text{ (Correct answer)}}$ However, because $x$ is odd, you likely entered it for the wrong integral bounds.

4. $\int_{0}^\infty \frac{e^x}{(6 + e^x)^2} \, dx$

Step 1: Substitution

Let $u = 6 + e^x$, so $du = e^x dx$. When:

• $x = 0$, $u = 6 + 1 = 7$,
• $x \to \infty$, $u \to \infty$.

The integral becomes: $\int_{7}^\infty \frac{1}{u^2} \, du.$

Step 2: Evaluate the integral

The antiderivative of $\frac{1}{u^2}$ is $-\frac{1}{u}$: $\left[ -\frac{1}{u} \right]_7^\infty = 0 - \left( -\frac{1}{7} \right) = \frac{1}{7}.$

Thus, the integral converges to: $\boxed{\frac{1}{7}}$

5. $\int_{-\infty}^\infty 5x e^{-x^2} \, dx$

Step 1: Analyze the integrand

The function $f(x) = 5x e^{-x^2}$ is odd, since $x$ is odd and $e^{-x^2}$ is even: $f(-x) = -f(x).$

For an odd function integrated over a symmetric interval $[-\infty, \infty]$, the integral is: $\boxed{0}$

Would you like detailed explanations for any of these? Let me know if you have questions!

Relative Questions:

1. Why do odd functions yield zero when integrated over symmetric intervals?
2. How does substitution simplify evaluating improper integrals?
3. What are the convergence conditions for integrals with exponential terms?
4. Can all improper integrals with infinity as bounds be simplified through substitution?
5. How do symmetry properties help simplify definite integrals?

Tip: When evaluating an integral, check the symmetry of the function first—it can save you significant time!