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Program of Lessons

# Integration methods

Integration by parts. Integration by substitution ( exchange ).

Integration by parts. If functions u ( x ) and v ( x ) have continuous first derivatives and the integral v ( x ) du ( x ) exists, then the integral u ( x ) dv ( x ) also exists and the equality

u ( x ) dv ( x ) = u ( x ) v ( x ) – v ( x ) du ( x )

takes place, or shortly:

u dv = u v v du .

Pay attention to the fact, that this operation and a differential of product of two functions are mutually inverse operations ( check, please ! ).

 E x a m p l e. Find the integral: ln x dx . S o l u t i o n. If to suppose that u = ln x and dv = dx, then du = dx / x and v = x . Using the formula of integration by parts, we receive:

Integration by substitution (exchange). If a function f ( z ) is given and has a primitive at z Z, a function z = g ( x ) has a continuous derivative at x X, and g ( X ) Z , then the function F ( x ) = f [ g ( x )] g' ( x ) has a primitive in Õ and

F ( x ) dx = f [ g ( x )] g' ( x ) dx = f ( z ) dz .

 E x a m p l e. Find the integral: . S o l u t i o n. To get rid of the square root we assume , then x = u2 + 3 and hence, dx = 2u du. Then exchanging, we have:

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