To evaluate the improper integral

$\int_0^{\infty} \frac{1}{(x+3)^2} \, dx,$

we can follow these steps:

1. Set up the integral with a limit to handle the improper upper limit:

   $\int_0^{\infty} \frac{1}{(x+3)^2} \, dx = \lim_{b \to \infty} \int_0^b \frac{1}{(x+3)^2} \, dx.$

2. Evaluate the integral: We can simplify this by using a substitution. Let $u = x + 3$. Then $du = dx$, and when $x = 0$, $u = 3$; when $x = b$, $u = b + 3$. The integral becomes

   $\lim_{b \to \infty} \int_3^{b+3} \frac{1}{u^2} \, du.$

3. Integrate $\frac{1}{u^2}$ with respect to $u$: The antiderivative of $\frac{1}{u^2}$ is $-\frac{1}{u}$, so

   $\lim_{b \to \infty} \left[ -\frac{1}{u} \right]_3^{b+3}.$

4. Evaluate the limits: Substituting the limits, we have

   $\lim_{b \to \infty} \left( -\frac{1}{b+3} + \frac{1}{3} \right).$

   As $b \to \infty$, the term $-\frac{1}{b+3}$ approaches 0. Therefore, we get

   $\frac{1}{3}.$

Thus, the value of the integral is

$\int_0^{\infty} \frac{1}{(x+3)^2} \, dx = \frac{1}{3}.$

Would you like more details or have any questions?

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Related Questions

1. How would the solution change if the integrand were $\frac{1}{(x+3)^3}$ instead?
2. Can we apply a different method, such as integration by parts, to solve this integral?
3. How would we solve $\int_0^{\infty} \frac{1}{(x+a)^2} \, dx$ for an arbitrary constant $a > 0$?
4. What if the integral was $\int_0^{\infty} \frac{1}{(x+3)^p} \, dx$ for some $p > 1$?
5. How does the behavior of the function $f(x) = \frac{1}{(x+3)^2}$ as $x \to \infty$ affect the convergence of the integral?

Tip

For improper integrals with infinite limits, substituting to shift the bounds and applying limits after integration often simplifies calculations.