Mathematics, 14.07.2021 14:00 shilohtito

What is the second integral of
?

$\frac{1}{x}$

Answers: 2

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Answers

Answer from: AdamFrost

$\displaystyle \int {\int {\frac{1}{x}} \, dx} \, dx = xln|x| - x + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Logarithmic Differentiation

Integration

IntegralsIndefinite IntegralsIntegration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Logarithmic Integration

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

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {\int {\frac{1}{x}} \, dx} \, dx$

Step 2: Integrate Pt. 1

[Inner Integral] Logarithmic Integration: $\displaystyle \int {ln|x|} \, dx$

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

Set u: $\displaystyle u = ln|x|$ [u] Differentiate: $\displaystyle du = \frac{1}{x} \ dx$ Set dv: $\displaystyle dv = dx$ [dv] Integrate: $\displaystyle v = x$

Step 4: Integrate Pt. 3

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

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

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

< 6 cm to > 26 cm i believe

step-by-step explanation:

Answer from: Quest

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ), by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i 0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5