THE METHOD OF INTEGRATION BY PARTS

All of the following problems use the method of integration by parts. This method uses the fact that the differential of function

is

.

For example, if

,

then the differential of is

.

Of course, we are free to use different letters for variables. For example, if

,

then the differential of is

.

When working with the method of integration by parts, the differential of a function will be given first, and the function from which it came must be determined. For example, if the differential is

,

then the function

leads to the correct differential. In general, function

,

where is any real constant, leads to the correct differential

.

When using the method of integration by parts, for convenience we will always choose when determining a function (We are really finding an antiderivative when we do this.) from a given differential. For example, if the differential of is

then the constant can be "ignored" and the function (antiderivative) can be chosen to be

.

The formula for the method of integration by parts is given by

.

This formula follows easily from the ordinary product rule and the method of u-substitution. Theoretically, if an integral is too "difficult" to do, applying the method of integration by parts will transform this integral (left-hand side of equation) into the difference of the product of two functions and a new ``easier" integral (right-hand side of equation). It is assumed that you are familiar with the following rules of differentiation.

  • a.)
  • b.)
  • c.)
  • d.)
  • e.)
  • f.)

We will assume knowledge of the following well-known, basic indefinite integral formulas :

  1. , where is a constant
  2. , where is a constant

Most of the following problems are average. A few are challenging. Make careful and precise use of the differential notation and and be careful when arithmetically and algebraically simplifying expressions

SOLUTIONS TO INTEGRATION BY PARTS

SOLUTION 1 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 2 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 3 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 4 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 5 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 6 : Integrate . Let

and

so that (Don't forget to use the chain rule when differentiating .)

and .

Therefore,

.

Now use u-substitution. Let

so that

,

or

.

Then

+ C

+ C

+ C .

SOLUTION 7 : Integrate . Let

and

so that

and .

Therefore,

.

SOLUTION 8 : Integrate . Let

and

so that

and .

Therefore,

(Add in the numerator. This will replicate the denominator and allow us to split the function into two parts.)

.

SOLUTION 9 : Integrate . Let

and

so that

and .

Therefore,

.

Integrate by parts again. Let

and

so that

and .

Hence,

.

1