web-1 - Review from Calculus I. In 133, Find each integral...

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Review from Calculus I. In 1–33, Find each integral by using one of the Basic 16 Integration Formulas. Do not use the substitution method. Simply recognize the Basic Integration Formula which applies and use it. When solving an integral problem such as these, the uppermost question in mind should be: This integral looks like it could be made to fit which formula?— Not: What shall I substitute? 1. Z (3 + x 2 ) 3 2 xdx = 1 2 Z (3 + x 2 ) 3 2 2 xdx = 1 5 (3 + x 2 ) 5 2 + C by B1 2. Z 5 - 2 x 2 xdx = - 1 4 Z (5 - 2 x 2 ) 1 2 ( - 4 xdx ) = - 1 6 (5 - 2 x 2 ) 3 2 + C by B1 3. Z (9 + 4 x + x 2 ) 2 3 (2 + x ) dx = 1 2 Z (9 + 4 x + x 2 ) 2 3 (4 + 2 x ) dx = 3 10 (9 + 4 x + x 2 ) 5 3 + C by B1 4. Z cos3 x dx 4 + sin3 x = 1 3 Z (4 + sin3 x ) - 1 2 3cos3 x dx =? by B1 5. Z 4 + e 3 x e 3 x dx = 1 3 Z (4 + e 3 x ) 1 2 3 e 3 x dx 6. Z (4 + ln x ) 5 x dx = Z (4 + ln x ) 5 1 x dx = 1 6 (4 + ln x ) 6 + C by B1 7. Z
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web-1 - Review from Calculus I. In 133, Find each integral...

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