What is the integral of the function y = 4x ln x?

What is the integral of the function y = 4x ln x?

• A. 2x² ln x - x² + C
• B. 4x² ln x - 2x² + C
• C. 2x ln x - x² + C
• D. 3x ln x + C

Answer: (A)

To calculate the integral of the function \( y = 4x \ln x \), we can utilize integration by parts, which is often useful for functions that are products of two differing types of functions, such as polynomial and logarithmic functions. In integration by parts, we use the formula: \[ \int u \, dv = uv - \int v \, du \] For the integral \( \int 4x \ln x \, dx \), we can choose: - \( u = \ln x \) (thus \( du = \frac{1}{x} \, dx \)) - \( dv = 4x \, dx \) (thus \( v = 2x^2 \)) Now, substitute these into the integration by parts formula: \[ \int 4x \ln x \, dx = 2x^2 \ln x - \int 2x^2 \cdot \frac{1}{x} \, dx \] Simplifying the integral on the right gives: \[ \int 2x^2 \cdot \frac{1}{x} \, dx = \int 2x \, dx = x^

To calculate the integral of the function ( y = 4x \ln x ), we can utilize integration by parts, which is often useful for functions that are products of two differing types of functions, such as polynomial and logarithmic functions.

In integration by parts, we use the formula:

[

\int u , dv = uv - \int v , du

]

For the integral ( \int 4x \ln x , dx ), we can choose:

• ( u = \ln x ) (thus ( du = \frac{1}{x} , dx ))

• ( dv = 4x , dx ) (thus ( v = 2x^2 ))

Now, substitute these into the integration by parts formula:

[

\int 4x \ln x , dx = 2x^2 \ln x - \int 2x^2 \cdot \frac{1}{x} , dx

]

Simplifying the integral on the right gives:

[

\int 2x^2 \cdot \frac{1}{x} , dx = \int 2x , dx = x^