Evaluate integral ln (3x) dx. Here we don't have much choice for u and dv. Let u = ln (3x) Then du = dx/x dx v = x Integrating by parts, we get integral ln (3x) dx = 3 ln (3x) -integral x middot dx/x = 3 ln (3x) – integral dx – x ln (3x) – x +
The Correct Answer and Explanation is :
To solve the integral (\int \ln(3x) \, dx) correctly using integration by parts, let’s clarify the steps.
The Correct Solution:
Using the formula for integration by parts:
[
\int u \, dv = uv – \int v \, du
]
Step 1: Choose (u = \ln(3x)) and (dv = dx).
Then, (du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} dx) (using the chain rule).
Also, integrating (dv = dx), we get (v = x).
Step 2: Apply the formula:
[
\int \ln(3x) \, dx = x \ln(3x) – \int x \cdot \frac{1}{x} \, dx
]
Step 3: Simplify the remaining integral:
The second term simplifies:
[
\int x \cdot \frac{1}{x} \, dx = \int dx = x
]
Thus, the integral becomes:
[
\int \ln(3x) \, dx = x \ln(3x) – x + C
]
Final Answer:
[
\int \ln(3x) \, dx = x \ln(3x) – x + C
]
Explanation (300 words):
The integral (\int \ln(3x) \, dx) is solved using the integration by parts technique, where the logarithmic function is a natural choice for (u), and (dx) becomes (dv). This choice works because the derivative of (\ln(3x)) simplifies to (\frac{1}{x}), making it manageable to integrate.
The formula for integration by parts is used to break the integral into two parts: (uv – \int v \, du). Here, (u = \ln(3x)), whose derivative is straightforward, and (dv = dx), whose integral is simply (v = x). Substituting these into the formula results in (x \ln(3x) – \int x \cdot \frac{1}{x} \, dx).
The remaining integral, (\int x \cdot \frac{1}{x} \, dx), simplifies to (\int dx = x). Combining this with the earlier term gives the final result:
[
\int \ln(3x) \, dx = x \ln(3x) – x + C.
]
The (C) represents the constant of integration, as this is an indefinite integral.
This result highlights the importance of choosing (u) and (dv) wisely in integration by parts. The logarithmic term (\ln(3x)) multiplies with the linear (x), but the extra term cancels out to produce a clean, simplified expression.