What term is subtracted from (x^4/4) ln(x) in the solution of the integral?

What term is subtracted from (x^4/4) ln(x) in the solution of the integral?

• A. (x^4/8)
• B. (x^4/16)
• C. (x^4/12)
• D. (x^4/10)

Answer: (B)

To determine the term that is subtracted from \(\frac{x^4}{4} \ln(x)\) in the integration process, we need to consider the integration by parts method. In this context, let's examine the integral \(\int x^4 \ln(x) \, dx\). Using integration by parts, we let: - \(u = \ln(x)\) which gives \(du = \frac{1}{x} \, dx\) - \(dv = x^4 \, dx\) which integrates to \(v = \frac{x^5}{5}\) According to the integration by parts formula, \(\int u \, dv = uv - \int v \, du\), we obtain: \[ \int x^4 \ln(x) \, dx = \frac{x^5}{5} \ln(x) - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx \] This simplifies the second integral to: \[ \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx = \int \frac{x^4}{5} \,

To determine the term that is subtracted from (\frac{x^4}{4} \ln(x)) in the integration process, we need to consider the integration by parts method. In this context, let's examine the integral (\int x^4 \ln(x) , dx).

Using integration by parts, we let:

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

• (dv = x^4 , dx) which integrates to (v = \frac{x^5}{5})

According to the integration by parts formula, (\int u , dv = uv - \int v , du), we obtain:

[

\int x^4 \ln(x) , dx = \frac{x^5}{5} \ln(x) - \int \frac{x^5}{5} \cdot \frac{1}{x} , dx

]

This simplifies the second integral to:

[

\int \frac{x^5}{5} \cdot \frac{1}{x} , dx = \int \frac{x^4}{5} ,