CPM Homework Help : CALC Problem 8-18

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Home > CALC > Chapter Ch8 > Lesson 8.1.2 > Problem 8-18

8-18.

No calculator! Evaluate the following integrals.

$\int \frac { 8 x ^ { 3 } - 1 } { 6 x ^ { 4 } - 3 x } d x$
Hint 1(a):
$U$ -substitution.
Hint 2(a):
$U = 6x^4 − 3x$

$\int _ { 2 } ^ { 5 } \pi ( ( x + 1 ) ^ { 2 } - 3 ^ { 2 } ) d x$
Hint (b):
Before you integrate, factor the π out of the integrand. You may (or may not) choose to expand $(x + 1)^²$ .

$\int ( 5 x - 2 + \frac { 5 } { x + 3 } ) d x$
Hint (c):
Recall that $\frac{5}{x+3}$ is a transformation of $\frac{1}{x}$ , and you know
the antiderivative of $\frac{1}{x}$ .

$\int 2 x \operatorname { sin } ( 11 x ^ { 2 } - 3 ) d x$
Hint (d):
$U$ -substitution.

$\int _ { - 1 } ^ { 0 } \frac { 2 x } { x + 2 } d x$
Hint (e):
You could long divide first or... use $U$ -Substitution.
Let $U =$ the denominator.
Steps (e):
If $U = x + 2$ then $x = U-2$ .
Therefore numerator
$2x=2(U-2) \text{ and }\frac{dU}{dx}=1 \text{ so }dU=dx.$
Also bounds $U(0) = 2$ and $U(− 1) = 1$
Rewrite the integral and evaluate.

$\int \frac { 3 } { \sqrt { 1 - 9 x ^ { 2 } } } d x$
Hint (f):
Recall that $9x^² = (3x)^²$
Then look for a familiar antiderivative.

Compute without a calculator

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