How do you compute the integral of ln(x) with respect to x?

How do you compute the integral of ln(x) with respect to x?

• A. xln(x) - x + C
• B. ln(x) + x + C
• C. xln(x) + x + C
• D. ln(x) - x + C

Answer: (A)

To compute the integral of \( \ln(x) \) with respect to \( x \), we can use integration by parts. Integration by parts is based on the formula: \[ \int u \, dv = uv - \int v \, du \] For the integral \( \int \ln(x) \, dx \), we can choose: - \( u = \ln(x) \) which gives \( du = \frac{1}{x} \, dx \) - \( dv = dx \) which gives \( v = x \) Now, applying the integration by parts formula: \[ \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx \] This simplifies to: \[ = x \ln(x) - \int 1 \, dx \] Calculating the remaining integral: \[ = x \ln(x) - x + C \] where \( C \) is the constant of integration. Therefore, the correct expression for the integral of \( \ln(x) \) with respect to \( x \) is: \[ x \ln(x) -

To compute the integral of ( \ln(x) ) with respect to ( x ), we can use integration by parts. Integration by parts is based on the formula:

[

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

]

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

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

• ( dv = dx ) which gives ( v = x )

Now, applying the integration by parts formula:

[

\int \ln(x) , dx = x \ln(x) - \int x \cdot \frac{1}{x} , dx

]

This simplifies to:

[

= x \ln(x) - \int 1 , dx

]

Calculating the remaining integral:

[

= x \ln(x) - x + C

]

where ( C ) is the constant of integration. Therefore, the correct expression for the integral of ( \ln(x) ) with respect to ( x ) is:

[

x \ln(x) -