

This is the missing power, n = -1, not covered by the power rule on
worksheet 8-1. Consider the integral as a fraction where the
denominator is to the first power and the numerator is the derivative
of the denominator [du/u]. Or the numerator can be the derivative of
the denominator except for a constant factor, which can be "fixed".
The absolute value is on the natural logarithm because logs are only
defined for positive values. The domain of the log x is x > 0.
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Problem 1:
![]() | The denominator is to the first power. Let the denominator = u = 2 + sin x. Then du = (0 + cos x) dx = cos x dx |
| This matches the numerator so it's |
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= ln |2 + sin x| + C = ln (2 + sin x) + C | |
NOTE: The answer to Example 1 above can be written without the absolute value because (2 + sin x) is always positive [sin x is always between -1 and +1, so (2 + sin x) is between +1 and +3].
Problem 2:
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let u = x³ + 2, then du = 3x² dx, "fix" the 3 |
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Problem 3:
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let u = 2x + 3, then du = 2 dx, "fix" the 2 |
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Problem 4:
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let u = tan x, then du = sec²x dx |
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Problem 5:
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Problem 6:
![]() | The denominator is NOT to the first power. This is not a log integral. The denominator is the 1/2 power so rewrite using exponents. |
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