Problem 11 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{\ln 2 x}{x} d x$$

Step-by-Step Solution

Answer

Using integration by parts, we have solved the given integral: \(\int \frac{\ln(2x)}{x} dx\) and found its solution to be: $$\int \frac{\ln(2x)}{x} dx = \ln(2x)\ln|x| - (\int \ln|x| \frac{1}{x} dx) + C$$ where C represents the constant of integration.

Step 1: Choose the functions u and dv

Choose the functions u and dv in order to apply integration by parts. Let u = \(\ln(2x)\) and dv = \(\frac{1}{x}dx\).

Step 2: Derive u and find v

Differentiate u with respect to x to find du: \(du = \frac{d(\ln(2x))}{dx} = \frac{2}{2x} dx = \frac{1}{x} dx\) Now, integrate dv with respect to x to find v: \(v = \int \frac{1}{x} dx = \ln|x| + C_1\)

Step 3: Apply the integration by parts formula

Integration by parts formula is: \(\int u dv = uv - \int v du\) Now substituting the u, v, and du into the formula, we get: \(\int \frac{\ln(2x)}{x} dx = \ln(2x)(\ln|x| + C_1) - \int (\ln|x| + C_1) \frac{1}{x} dx\)

Step 4: Solve the remaining integral

Split the integral into two parts and solve: \(= \ln(2x)(\ln|x| + C_1) - ( \int \ln|x| \frac{1}{x} dx + \int \frac{C_1}{x} dx)\) \(= \ln(2x)(\ln|x| + C_1) - [( \int \ln|x| \frac{1}{x} dx) + ( C_1 \int \frac{1}{x}dx)]\) Note that the second integral of the right-hand side has already been solved as \(v\): \(= \ln(2x)(\ln|x| + C_1) - (\int \ln|x| \frac{1}{x} dx) - (C_1 (\ln|x| + C_2))\)

Step 5: Simplify the expression and combine constants

Combining the constants in the expression, we get: \(\int \frac{\ln(2x)}{x} dx = \ln(2x)(\ln|x|) - (\int \ln|x| \frac{1}{x} dx) + C\) The solution to the integral is: $$\int \frac{\ln(2x)}{x} dx = \ln(2x)\ln|x| - (\int \ln|x| \frac{1}{x} dx) + C$$