Antiderivative of f(x) = x · ln(2x)

This problem asks to find the antiderivative F(x) of the function f(x) = x · ln(2x) satisfying the condition F(1) = 0.

Problem

$f (x) = x \cdot ln (2 x), F (1) = 0$

Solution

Step 1

Apply the integration by parts formula:

\int u d v = u v - \int v d u

Step 2

Choose u = ln(2x) (then du = (1/x) dx) and dv = x dx (then v = x²/2):

u = ln (2 x), d u = \frac{1}{x} d x

d v = x d x, v = \frac{x^{2}}{2}

Step 3

Substitute into the integration by parts formula:

\int x \cdot ln (2 x) d x = \frac{x^{2} \cdot ln (2 x)}{2} - \int \frac{x^{2}}{2 x} d x

Step 4

Evaluate the remaining integral and obtain the general expression for the antiderivative:

F (x) = \frac{x^{2} \cdot ln (2 x)}{2} - \frac{x^{2}}{4} + C

Step 5

Find the constant C from the condition F(1) = 0. Note that ln 2 is a constant:

F (1) = \frac{ln 2}{2} - \frac{1}{4} + C = 0

Step 6

From this, C = 1/4 − (1/2)·ln 2. The required antiderivative has the form:

Answer

$F (x) = \frac{x^{2} \cdot ln (2 x)}{2} - \frac{x^{2}}{4} + \frac{1}{4} - \frac{ln 2}{2}$

Frequently asked questions

How do I find the antiderivative of f(x) = x · ln(2x)?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = ln(2x) (the logarithm simplifies when differentiated) and dv = x dx (easily integrated to x²/2).

What is the derivative of ln(2x)?

By the chain rule: (ln(2x))′ = 1/(2x) · 2 = 1/x. The factor 2 from the argument cancels with the denominator 2x. Hence du = (1/x) dx — the same as for ln x.

Why does C contain ln 2?

From the condition F(1) = 0: F(1) = ln(2·1)/2 − 1/4 + C = ln 2/2 − 1/4 + C = 0. Hence C = 1/4 − ln 2/2. The value ln 2 does not cancel because the upper limit is 1, not 1/2.

When is integration by parts used for logarithms?

When the integrand contains a logarithm in a product with a polynomial or by itself. In such cases, take the logarithm as u and the rest as dv. After differentiating the logarithm, a simple rational function appears, and the integral reduces to a tabular one.