Question

What is the second integral of

`(1)/(x)`
?​

Answer 1

`\displaystyle \int {\int {(1)/(x)} \, dx} \, dx = xln|x| - x + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Logarithmic Differentiation

Integration

• Integrals
• Indefinite Integrals
• Integration Constant C

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

Logarithmic Integration

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 {\int {(1)/(x)} \, dx} \, dx`

Step 2: Integrate Pt. 1

1. [Inner Integral] Logarithmic Integration:
   `\displaystyle \int ln \, dx`

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

1. Set u:
   `\displaystyle u = ln|x|`
2. [u] Differentiate:
   `\displaystyle du = (1)/(x) \ dx`
3. Set dv:
   `\displaystyle dv = dx`
4. [dv] Integrate:
   `\displaystyle v = x`

Step 4: Integrate Pt. 3

1. [Integral] Integration by Parts:
   `\displaystyle \int x \, dx = xln|x| - \int {(x \cdot (1)/(x))} \, dx`
2. [Right Integral] Simplify:
   `\displaystyle \int x \, dx = xln|x| - \int {} \, dx`
3. [Right Integral] Reverse Power Rule:
   `\displaystyle \int ln \, dx = xln|x| - x + C`
4. Redefine:
   `\displaystyle \int {\int {(1)/(x)} \, dx} \, dx = xln|x| - x + C`

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

Unit: Integration

Book: College Calculus 10e