14. Properties of the indefinite integral. Integration by substitution .Integration by parts.

Answer:

Suppose that F(x) and G(x) are antiderivativesof f(x) and g(x) respectively , and that C=const. then:

1. A constant factor can be moved through an integral sign ; that is

f(x)dx=c (x)dx=cF(x) + C.

2. An antiderivative of a sum is the sum of the antiderivatives, that is:

= + =F(x) + G(x) + C.

3. An antiderivative of a difference is the difference of the antiderivatives, that is:

= - = F(x) – G(x) + C.

Integration by substitution.

= F(g(x)) + C. (1)

(1) can be expressed as: = U + C. (2)

The process of evaluating an integral of form (1) by converting it into (2) with the substitution :

U = g(x) and dU = g’(x)dx

is called the method of U-substitution.

Integration by parts:

15.Integrating rational function. Integrating Binomial differentials. Euler’s substitution.

Answer: Integration rational function: (1)

is called a proper rational function, if the degree od the numerator is less than the degree of the denominator (n<k)

There is a theorem in advanced algebra which states that every proper rational function can be expressed as a sum:

= F₁(x) + F₂(x) + …+ F_n(x) (2) – is called the rational fraction.

Where F₁(x), F₂(x),…. are rational function of the form:

A/(ax+b)^k or Ax+b/(ax²+bx+c)^k

There are 4 type of partial fractions.

I type: A/x-x₀ (the denominator has a real root)

II type: A/(x-x₀)^α (α>1, the denominator has a real multiple root)

III type: Ax+b/x²+bx+c(the denominator has a complex root )

IV type: Ax+b/(ax²+bx+c)^β (β>1, the denominator has a complex multiple root).

If in (1) n k, the (1) is called improper rational function:

= M(x) + , - improper rational function, M(x) – integer part, - proper rational function. Integrating Binomial differentials: ^m(a+bxⁿ)^pdx - Binomial differ.

1. p Z, then x= t^N where N is common denominator of m and n.

2. Z, then a+bxⁿ=t^N is denominator of p.

3. +p Z , then a+bxⁿ/xⁿ=t^N, where N is denominator of p.

Euler’s substitutions: R(x )

1. a>0 => ) = x + t

2. c>0 => )= + xt

3. = t(x-x₁)