There is only one difference between indefinite integral method of substitution and definite integral method of substitution: indefinite integral method of substitution must be changed back to the original variable in the end, and the upper and lower limits of definite integral method of substitution must be changed accordingly, and it is not necessary to be changed back to the original variable in the end.

Solution of indefinite integral method of substitution;

Let g be a differentiable function and function f be its derivative.

Then ∫ f (g (x)) g' (x) = f (g (x))+C. Let u=g(x), then du=g'(x)dx,?

Then ∫ f (g (x)) g' (x) = ∫ f (u) du = f (u)+c = f (g (x))+c.

Substitution means changing the integral variable to u=g(x).

Replacement method of definite integral;

Let the function f(x) be continuous in the interval [a, b]; The function g(t) is single-valued and has a continuous derivative in the interval [m, n]; When t changes in the interval [m, n], the value of x=g(t) changes in [a, b], g (m) = a, g (n) = b; Then there is the substitution formula of definite integral:

Extended data:

Solutions other than indefinite integral substitution method and definite integral substitution method;

Let the function and U, V have continuous derivatives, then d(uv)=udv+vdu. Pass the term to get udv=d(uv)-vdu.

Integral on both sides, and get the partial integral formula.

∫udv=uv-∫vdu .？ ⑴

The formula (1) is called the partial integral formula. If the integral ∫vdu is easy to find, you will get the left-handed integral formula.

The key to the application of partial integral formula is to choose u and v appropriately.

Generally speaking, u, v? The selection principle is:

1, choose v for integration and u for simple derivation.

Example: u = inx and v = x should be set at ∫Inx dx.

The essence of partial integration is to change the required integral into the difference between two integrals, and integrate easily first. It's actually two integrals.

Rational functions can be divided into algebraic expressions (polynomials) and fractions (quotients of two polynomials). Fractions can be divided into true fractions and false fractions, and can be converted into the sum of an algebraic expression and a true fraction by polynomial division. Therefore, this problem can be transformed into calculating the integral of the true fraction.

It can be proved that any proper fraction can always be decomposed into the sum of partial fractions.

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