Question

Integrate by parts logx.dx​

Answer 1

`\displaystyle \int {\log x} \, dx = x \bigg( \log x - (1)/(\ln(10)) \bigg) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Integration

• Integrals
• [Indefinite Integrals] Integration Constant C

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

Integration by Parts:
`\displaystyle \int {u} \, dv = uv - \int {v} \, du`

• [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Explanation:

Step 1: Define

Identify

`\displaystyle \int {\log x} \, dx`

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

1. Set u:
   `\displaystyle u = \log x`
2. [u] Logarithmic Differentiation:
   `\displaystyle du = (1)/(x \ln(10)) \ dx`
3. Set dv:
   `\displaystyle dv = dx`
4. [dv] Integration Rule [Reverse Power Rule]:
   `\displaystyle v = x`

Step 3: integrate Pt. 2

1. [Integral] Integration by Parts:
   `\displaystyle \int {\log x} \, dx = x \log x - \int {(1)/(\ln(10))} \, dx`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {\log x} \, dx = x \log x - (1)/(\ln(10)) \int {} \, dx`
3. [Integral] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int {\log x} \, dx = x \log x - (1)/(\ln(10))x + C`
4. Factor:
   `\displaystyle \int {\log x} \, dx = x \bigg( \log x - (1)/(\ln(10)) \bigg) + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration