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.

**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 (du/u) = ln |u| + C | |

= 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:**

let u = x³ + 2, then du = 3x² dx, "fix" the 3 | |

**Problem 3:**

** **

let u = 2x + 3, then du = 2 dx, "fix" the 2 | |

**Problem 4:**

let u = tan x, then du = sec²x dx | |

**Problem 5:**

**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. | |