Integrating polynomial functions

EXAMPLE - Rewriting the Integrand

Before beginning the exercise set, be sure you realize that one of the most important steps in finding antiderivatives is rewriting the integrand in a form that fits the basic integration rules. Look at the several aditional examples.

Given integral

\to

Rewrite

\to

Integrate

\to

Simplify

This pattern is followed in the next examples.

∫ 2 x d x = 2 ∫ x - 1 / 2 d x =

= 2 ( x 1 / 2 1 / 2 ) + C = 4 x 1 / 2 + C

∫ ( x 2 + 1 ) 2 d x = ∫ ( x 4 + 2 x 2 + 1 ) d x =

= x 5 5 + 2 ( x 3 3 ) + x + C = x 5 5 + 2 x 3 3 + x + C

∫ x 3 + 3 x 2 d x = ∫ ( x + 3 x - 2 ) d x =

= x 2 2 + 3 ( x - 1 - 1 ) + C = x 2 2 - 3 x + C

∫ x 3 ⁢ ( x - 4 ) ⅆ x = ∫ ( x 4 / 3 - 4 x 1 / 3 ) d x =

= x 7 / 3 7 / 3 - 4 ( x 4 / 3 4 / 3 ) + C = 3 x 4 / 3 7 ( x - 7 ) + C

In Exercises 1 - 6, complete the table using this example as a model

	Given integral	Rewrite	Integrate	Simplify
1.	∫ u ⋅ ( 3 u 2 + 1 ) d u
2.	∫ x ( x + 1 ) d x
3.	∫ ( x + 1 ) ( 3 x - 2 ) d x
4.	∫ ( 2 t 2 - 1 ) 2 d t
5.	∫ ( 1 + 3 t ) t 2 d t
6.	∫ y 2 y d y

	∫ 2 x d x	= 2 ∫ x - 1 / 2 d x =
		= 2 ( x 1 / 2 1 / 2 ) + C = 4 x 1 / 2 + C

	∫ ( x 2 + 1 ) 2 d x	= ∫ ( x 4 + 2 x 2 + 1 ) d x =
		= x 5 5 + 2 ( x 3 3 ) + x + C = x 5 5 + 2 x 3 3 + x + C

	∫ x 3 + 3 x 2 d x	= ∫ ( x + 3 x - 2 ) d x =
		= x 2 2 + 3 ( x - 1 - 1 ) + C = x 2 2 - 3 x + C

	∫ x 3 ⁢ ( x - 4 ) ⅆ x	= ∫ ( x 4 / 3 - 4 x 1 / 3 ) d x =
		= x 7 / 3 7 / 3 - 4 ( x 4 / 3 4 / 3 ) + C = 3 x 4 / 3 7 ( x - 7 ) + C