Chapter 5 | Integration
611
Example 5.52 Finding an Antiderivative Involving the Inverse Tangent Function Find an antiderivative of ⌠ ⌡ 1 1+4 x 2 dx .
Solution Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for tan −1 u + C . So we use substitution, letting u =2 x , then du =2 dx and 1/2 du = dx . Then, we have
⌠ ⌡
1 2
1 1+ u 2
−1 u + C = 1
−1 (2 x )+ C .
du = 1
2 tan
2 tan
Use substitution to find the antiderivative ⌠ ⌡
5.42
dx 25+4 x 2 .
Example 5.53 Applying the Integration Formulas Find the antiderivative of ⌠ ⌡ 1 9+ x 2 dx .
Solution Apply the formula with a =3. Then,
⌠ ⌡
−1 ⎛
⎞ ⎠ + C .
dx 9+ x 2
⎝ x 3
= 1 3 tan
Find the antiderivative of ⌠ ⌡
5.43
dx 16+ x 2
.
Example 5.54 Evaluating a Definite Integral Evaluate the definite integral ⌠ ⌡ 3/3 3 dx 1+ x 2 .
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