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8-42.

No calculator! Evaluate the integrals below.

$\int _ { 2 } ^ { 4 } \frac { 1 } { 9 - 2 x } d x$
Step 1(a):
Before integrating, make a choice:
You could factor out $\frac{1}{2}$ or you could let U = 9 − 2x.
Hint 1(a):
$\int \frac{1}{x}dx=\text{ln}\left | x \right |+C$
Don't forget the absolute value!
Hint 2(a):
$\text{ln}|a|-\text{ln}|b|=\text{ln}\left | \frac{a}{b} \right |$

$\int _ { - 3 } ^ { 3 } ( 2 x ^ { 5 } + 3 x ^ { 2 } + 1 ) d x$
Hint (b):
Notice the symmetrical bounds.Is this function odd? If so, the area will be 0. If not, integrate.

$\int _ { 3 } ^ { 2 } \frac { x ^ { 2 } - 1 } { x - 1 } d x$
Hint (c):
Factor before integrating.

$\int _ { 1 } ^ { 4 } \sqrt { 16 x ^ { 3 } } d x$
Hint (d):
Recall that $\sqrt{16x^{3}}=4x^{\frac{3}{2}}$ .

$\int _ { \pi / 4 } ^ { \pi / 2 } \operatorname { sin } ^ { 3 } x \operatorname { cos } x d x$
Hint (e):
U-substitution. Should U = sinx or U = cosx?

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