Mathematics, 30.01.2020 14:59 braydenmcd02

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find the indefinite integral

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Answer from: gabriellebarnes66

$\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = 2arctan(x) + \frac{ln|x^2 + 1|}{2} + C$

General Formulas and Concepts:

Algebra I

Terms/CoefficientsExpanding

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

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

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx$

Step 2: Integrate Pt. 1

Set variables for u-substitution.

Set u: $\displaystyle u = x^2 + 1$ [u] Differentiate [Basic Power Rule, Addition/Subtraction]: $\displaystyle du = 2x \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2}\int {\frac{2(x + 2)}{x^2 + 1}} \, dx$ [Integrand] Expand: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2}\int {\frac{2x + 4}{x^2 + 1}} \, dx$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ \int {\frac{2x}{x^2 + 1}} \, dx + \int {\frac{4}{x^2 + 1}} \, dx \bigg]$ [2nd Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ \int {\frac{2x}{x^2 + 1}} \, dx + 4\int {\frac{1}{x^2 + 1}} \, dx \bigg]$ [1st Integral] U-Substitution: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ \int {\frac{1}{u}} \, du + 4\int {\frac{1}{x^2 + 1}} \, dx \bigg]$ [1st Integral] Logarithmic Integration: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ ln|u| + 4\int {\frac{1}{x^2 + 1}} \, dx \bigg]$ [Integral] Arctrig Integration: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ ln|u| + 4 \bigg( \frac{1}{1}arctan \Big( \frac{x}{1} \Big) \bigg) \bigg] + C$ Simplify: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = \frac{1}{2} \bigg[ ln|u| + 4arctan(x) \bigg] + C$ Expand: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = 2arctan(x) + \frac{ln|u|}{2} + C$ Back-Substitute: $\displaystyle \int {\frac{x + 2}{x^2 + 1}} \, dx = 2arctan(x) + \frac{ln|x^2 + 1|}{2} + C$

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

Unit: Integration

Answer from: Quest

table a is wierd

step-by-step explanation:

6*f

Answer from: Quest

Looks pretty good to me!

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