Evaluate the integral: $\int x \ln (x-8) d x$
To evaluate the indefinite integral \(\int x \ln(x-8) \, dx\), we identify the integrand as a product of two functions, \(f(x) = x\) and \(g(x) = \ln(x-8)\). The function is defined for \(x > 8\). We will use the integration by parts formula, which states that \(\int u \, dv = uv - \int v \, du\).
We choose \(u = \ln(x-8)\) and \(dv = x \, dx\).
Differentiating \(u\) gives \(du = \frac{1}{x-8} \, dx\).
Integrating \(dv\) gives \(v = \int x \, dx = \frac{1}{2}x^2\).
Applying the integration by parts formula:
\(\int x \ln(x-8) \, dx = \left(\frac{1}{2}x^2\right) \ln(x-8) - \int \left(\frac{1}{2}x^2\right) \frac{1}{x-8} \, dx\).
We need to simplify the integral \(\int \frac{x^2}{2(x-8)} \, dx\). We can rewrite the integrand by polynomial division:
\(\frac{x^2}{x-8} = \frac{x^2 - 64 + 64}{x-8} = \frac{(x-8)(x+8) + 64}{x-8} = x + 8 + \frac{64}{x-8}\).
Thus, the integral becomes:
\(\int \frac{x^2}{2(x-8)} \, dx = \frac{1}{2} \int \left(x + 8 + \frac{64}{x-8}\right) \, dx\).
Evaluating this term by term gives:
\(\frac{1}{2} \left(\frac{1}{2}x^2 + 8x + 64 \ln(x-8)\right) = \frac{1}{4}x^2 + 4x + 32 \ln(x-8)\).
Combining the parts together, we get:
\(\int x \ln(x-8) \, dx = \frac{1}{2}x^2 \ln(x-8) - \left(\frac{1}{4}x^2 + 4x + 32 \ln(x-8)\right) + C\).
Grouping the \(\ln(x-8)\) terms leads to:
\(\left(\frac{1}{2}x^2 - 32\right) \ln(x-8) - \frac{1}{4}x^2 - 4x + C\).
This is the final form of the indefinite integral.
Integration by parts is essential in fields like physics and engineering, where you often need to calculate work done by variable forces or find centers of mass for objects with varying densities. By breaking down complex integral expressions involving products of functions, you can derive accurate models to optimize performance in mechanical systems or electrical circuits. For instance, determining the energy distribution in a system often requires evaluating integrals of the form found in this problem.