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:
ln
x dx = x ln x –
x dx /
x = x ln x – x + C . |
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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