Mathematics, 23.10.2020 16:00 maysahdabest

Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) 1 /x^2 − 49 dx

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Answer from: jessicad1198

$\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Pre-Calculus

Partial Fraction Decomposition

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 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

$\displaystyle \int {\frac{1}{x^2 - 49}} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Factor: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\frac{1}{(x - 7)(x + 7)}} \, dx$ [Integrand] Split [Partial Fraction Decomp]: $\displaystyle \frac{1}{(x - 7)(x + 7)} = \frac{A}{x - 7} + \frac{B}{x + 7}$ Rewrite: $\displaystyle 1 = A(x + 7) + B(x - 7)$ [Decomp] Substitute in x = 7: $\displaystyle 1 = A(7 + 7) + B(7 - 7)$ Simplify: $\displaystyle 1 = 14A$ Solve: $\displaystyle A = \frac{1}{14}$ [Decomp] Substitute in x = -7: $\displaystyle 1 = A(-7 + 7) + B(-7 - 7)$ Simplify: $\displaystyle 1 = -14B$ Solve: $\displaystyle B = \frac{-1}{14}$ [Split Integrand] Substitute in variables: $\displaystyle \frac{1}{(x - 7)(x + 7)} = \frac{\frac{1}{14}}{x - 7} - \frac{\frac{1}{14}}{x + 7}$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Split Integrand]: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\bigg( \frac{\frac{1}{14}}{x - 7} - \frac{\frac{1}{14}}{x + 7} \bigg)} \, dx$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \int {\frac{\frac{1}{14}}{x - 7}} \, dx - \int {\frac{\frac{1}{14}}{x + 7}} \, dx$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14}\int {\frac{1}{x - 7}} \, dx - \frac{1}{14}\int {\frac{1}{x + 7}} \, dx$ Factor: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \int {\frac{1}{x - 7}} \, dx - \int {\frac{1}{x + 7}} \, dx \bigg)$

Step 4: Integrate Pt. 3

Identify variables for u-substitution.

Integral 1

Set u: $\displaystyle u = x - 7$ [u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = dx$

Integral 2

Set z: $\displaystyle z = x + 7$ [z] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle dz = dx$

Step 5: Integrate Pt. 4

[Integrals] U-Substitution: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \int {\frac{1}{u}} \, du - \int {\frac{1}{z}} \, dz \bigg)$ [Integrals] Logarithmic Integration: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |u| - \ln |z| \bigg) + C$ [Variables] Back-Substitute: $\displaystyle \int {\frac{1}{x^2 - 49}} \, dx = \frac{1}{14} \bigg( \ln |x - 7| - \ln |x + 7| \bigg) + C$

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

Unit: Integration

Answer from: Quest

the required function is $y=-x+4$ .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-7}{)}=\frac{-3}{3}=-1$

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

$y=mx+b$

where, m is slope and b is y-intercept.

$y=-1x+4$

$y=-x+4$

therefore the required function is $y=-x+4$ . the graph of this function is shown below.

Answer from: Quest

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes

Find the composition of transformations that map abcd to ehgf. !

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