Integrate indefinite integral:

Solution:
1. use substitution
=>

=>

=>

=>


2. decompose into partial fractions


where A=1/2, B=1/2, C=-1

3. Substitute partial fractions and continue


4. back-substitute u=e^x


Note: log(x) stands for natural log, and NOT log10(x)
Answer:
1/5
Step-by-step explanation:
1/9=0.11
1/3=0.33
1/5=0.20
I don't really know the answer but I tried.
Answer:
x=4
Step-by-step explanation:
11x-4=5x+20
4x=16
x=4
<h3>Answer:</h3>
- the result in scientific notation is 6.561×10⁻⁷
- the result as a decimal is 0.000 000 656 1
<h3>Explanation:</h3>
The rules of exponents say the exponent outside parentheses applies to each factor inside parentheses.
... (8.1*10^-4)^2 = 8.1^2 × (10^-4)^2
... = 65.61 × 10^-8
... = 6.561 × 10^-7 . . . . adjust to scientific notation with one digit left of the decimal point
The exponent of -7 means the decimal point in the decimal number is 7 places to the left of where it is in scientific notation. That is ...
... 6.561 × 10^-7 = 0.0000006561
13, 21, 34 are the next three numbers