The integral provided is:

$\int_{-\infty}^0 \frac{1}{(x - 5)^2} \, dx$

This integral appears to be improper due to the singularity at $x = 5$. Let us analyze and solve it step by step.

Step 1: Domain of Integration

• The singularity $x = 5$ does not lie within the limits of integration ($-\infty$ to $0$), so there is no singularity affecting this integral.
• The function $\frac{1}{(x - 5)^2}$ is well-defined and continuous on $(-\infty, 0)$.

Step 2: General Antiderivative

The integrand can be rewritten as: $\frac{1}{(x - 5)^2}$ Using the formula: $\int \frac{1}{(x - 5)^2} \, dx = -\frac{1}{x - 5} + C$

Step 3: Evaluate the Definite Integral

The definite integral from $-\infty$ to $0$ becomes: $\int_{-\infty}^0 \frac{1}{(x - 5)^2} \, dx = \left[-\frac{1}{x - 5}\right]_{-\infty}^0$

1. At the upper limit $x = 0$: $-\frac{1}{x - 5} \Big|_{x=0} = -\frac{1}{0 - 5} = \frac{1}{5}$

2. At the lower limit $x \to -\infty$: $-\frac{1}{x - 5} \Big|_{x \to -\infty} \to 0 \quad \text{(since as } x \to -\infty, x - 5 \to -\infty \text{)}.$

Final Result:

$\int_{-\infty}^0 \frac{1}{(x - 5)^2} \, dx = \frac{1}{5} - 0 = \frac{1}{5}$

Would you like me to break down any part of this further or provide additional examples?

5 Related Questions:

1. How do improper integrals differ from regular definite integrals?
2. What happens if the singularity lies within the limits of integration?
3. How do we determine convergence of improper integrals?
4. What are the implications of integrating $\frac{1}{x^2}$ over symmetric limits around zero?
5. Can we generalize the formula for integrating $\frac{1}{(x - a)^n}$?

Tip:

When evaluating improper integrals, always check for singularities or infinite limits carefully to ensure proper convergence analysis.