121616949-math.188 - arctan ± 2 √ 3 x 1 √ 3 C The...

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174 Chapter 8 Techniques of Integration Now we can substitute in the original integral, using x = ( 3 / 2)(tan u - 1 / 3) in the numerator. Z x x 2 + x + 1 dx = 4 3 Z ( 3 / 2)(tan u - 1 / 3) tan 2 u + 1 3 2 sec 2 u du = 4 3 Z ( 3 / 2)(tan u - 1 / 3) sec 2 u 3 2 sec 2 u du = 4 3 Z 3 2 tan u - 1 3 3 2 du = Z tan u - 1 3 du = - ln | cos u | - u 3 + C. = ln | sec u | - u 3 + C. Finally, we can substitute u = arctan 2 3 x + 1 3 and sec u = p tan 2 u + 1 = s 2 3 x + 1 3 2 + 1 to get ln fl fl fl fl fl fl s 2 3 x + 1 3 2 + 1 fl fl fl fl fl fl - 1 3
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Unformatted text preview: arctan ± 2 √ 3 x + 1 √ 3 ¶ + C. The details here are admittedly a bit unpleasant, but the whole process is fairly me-chanical and “easy” in principle. Exercises Find the antiderivatives. 1. Z 1 4-x 2 dx ⇒ 2. Z x 4 4-x 2 dx ⇒ 3. Z 1 x 2 + 10 x + 25 dx ⇒ 4. Z x 2 4-x 2 dx ⇒...
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