Answer:
![\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7B%5Cfrac%7B1%7D%7B9x%5E2%2B4%7D%7D%20%5C%2C%20dx%20%3D%20%5Cfrac%7B1%7D%7B6%7Darctan%28%5Cfrac%7B3x%7D%7B2%7D%29%20%2B%20C)
General Formulas and Concepts:
<u>Algebra I</u>
<u>Calculus</u>
Antiderivatives - integrals/Integration
Integration Constant C
U-Substitution
Integration Property [Multiplied Constant]: ![\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7Bcf%28x%29%7D%20%5C%2C%20dx%20%3D%20c%20%5Cint%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Trig Integration:
Step-by-step explanation:
<u>Step 1: Define</u>
<u />
<u />
<u />
<u>Step 2: Integrate Pt. 1</u>
- [Integral] Factor fraction denominator:
![\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7B%5Cfrac%7B1%7D%7B9%28x%5E2%20%2B%20%5Cfrac%7B4%7D%7B9%7D%29%7D%7D%20%5C%2C%20dx)
- [Integral] Integration Property - Multiplied Constant:
![\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B9%7D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bx%5E2%20%2B%20%5Cfrac%7B4%7D%7B9%7D%7D%7D%20%5C%2C%20dx)
<u>Step 3: Identify Variables</u>
<em>Set up u-substitution for the arctan trig integration.</em>
![\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20u%20%3D%20x%20%5C%5C%20a%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5C%5C%20du%20%3D%20dx)
<u>Step 4: Integrate Pt. 2</u>
- [Integral] Substitute u-du:
![\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B9%7D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bu%5E2%20%2B%20%28%5Cfrac%7B2%7D%7B3%7D%29%5E2%7D%20%5C%2C%20du)
- [Integral] Trig Integration:
![\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B9%7D%5B%5Cfrac%7B1%7D%7B%5Cfrac%7B2%7D%7B3%7D%7Darctan%28%5Cfrac%7Bu%7D%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%5D%20%2B%20C)
- [Integral] Simplify:
![\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B9%7D%5B%5Cfrac%7B3%7D%7B2%7Darctan%28%5Cfrac%7B3u%7D%7B2%7D%29%5D%20%2B%20C)
- [integral] Multiply:
![\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B6%7Darctan%28%5Cfrac%7B3u%7D%7B2%7D%29%20%2B%20C)
- [Integral] Back-Substitute:
![\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B6%7Darctan%28%5Cfrac%7B3x%7D%7B2%7D%29%20%2B%20C)
Topic: AP Calculus AB
Unit: Integrals - Arctrig
Book: College Calculus 10e
Answer:
1.234 × 101
Step-by-step explanation:
Answer: nth=4x
Step-by-step explanation: 4,8,16,32
So when x=1 nth=4
Then 4x should be the answer we need to find any term
He actually borrowed P=21349-3000=18349 (present value)
Assume the monthly interest is i.
then future value due to loan:
F1=P(1+i)^n=18349(1+i)^(5*12)=18349(1+i)^60
future value from monthly payment of A=352
F2=A((1+i)^n-1)/i=352((1+i)^60-1)/i
Since F1=F2 for the same loan, we have
18349(1+i)^60=352((1+i)^60-1)/i
Simplify notation by defining R=1+i, then
18349(R^60)-352(R^60-1)/(R-1)=0
Simplify further by multiplication by (R-1)
f(R)=18349*R^60*(R-1)-352(R^60-1)=0
Solve for R by trial and error, or by iteration to get R=1.004732
The APR is therefore
12*(1.004732-1)=0.056784, or 5.678% approx.
Six is the base because 6^4 is 1296.