What is the difference between a prime number and a complex number?

2 answers:

katrin2010 [14]2 years ago

7 0

A Prime number is a number is a number which cannot be divided by any numbers except for itself and one to give an integer

A Complex Number is any number containing Real and Imaginary parts ( $Imaginary= \sqrt{-1}=i$

mixer [17]2 years ago

6 0

"Prime"and "complex"are not opposites. If a number is a prime, then it has
exactly two factors ... ' 1 ' and itself. If it has any more factors, then it's not
a prime number, it's a "COMPOSITE" number.

(If a number is not a 'complex' number, then it's a REAL number.)

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Perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integra

worty [1.4K]

Answer:

$x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C$

Step-by-step explanation:

Given the integrand $\int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx$ , before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

Hence the partial form of the function $\dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}$

Integrating its partial sum

$\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx = \int\limits (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2} \int\limits{\frac{1}{x} } \, dx + \frac{2}{13} \int\limits{\frac{1}{x-2} } \, dx$