Integration methods
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
takes place, or shortly:
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
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 = u
^{
2
}
+ 3
and hence,
dx =
2
u du
. Then exchanging, we have:
