Find the integral of (log(x)) dx.

Find the integral of (log(x)) dx.

• A. x log(x) - x + C
• B. log(x^2)/2 + C
• C. x^2 log(x) + C
• D. log(x) + C

Answer: (A)

To find the integral of \( \log(x) \, dx \), we can use integration by parts. The integration by parts formula states that: \[ \int u \, dv = uv - \int v \, du \] For this integral, we can choose: - \( u = \log(x) \) (which means \( du = \frac{1}{x} \, dx \)), - \( dv = dx \) (which gives \( v = x \)). Applying the integration by parts formula, we have: \[ \int \log(x) \, dx = x \log(x) - \int x \cdot \frac{1}{x} \, dx \] Simplifying the second integral: \[ \int x \cdot \frac{1}{x} \, dx = \int 1 \, dx = x \] Now substituting this back into our equation gives: \[ \int \log(x) \, dx = x \log(x) - x + C \] Thus, the result of the integral is: \[ x \log(x) - x + C \] This confirms that the first choice

To find the integral of ( \log(x) , dx ), we can use integration by parts. The integration by parts formula states that:

[

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

]

For this integral, we can choose:

• ( u = \log(x) ) (which means ( du = \frac{1}{x} , dx )),

• ( dv = dx ) (which gives ( v = x )).

Applying the integration by parts formula, we have:

[

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

]

Simplifying the second integral:

[

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

]

Now substituting this back into our equation gives:

[

\int \log(x) , dx = x \log(x) - x + C

]

Thus, the result of the integral is:

[

x \log(x) - x + C

]

This confirms that the first choice