The reciprocal integral

Where it comes from

The problem: the power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C works for every exponent — except one. Plug in n = −1 and you get x&sup0;/0: division by zero, meaningless. So the innocent-looking integrand 1/x = x⁻¹ falls through a hole in the power rule, and something genuinely different has to fill it.

What function has derivative 1/x? For x > 0, the answer is ln x. But 1/x is also perfectly defined for negative x — and there ln x does not even exist. The tempting half-answer:

Before reading on: 1/x is perfectly defined at x = −3, but ln(−3) does not exist. The guess ln x + C therefore can’t be the whole story. What is the smallest fix that makes it work for negative x too?

Test it where it hurts: ∫₋₂⁻¹ dx/x. The guess says [ln x]₋₂⁻¹ — but ln(−2) is undefined, so the guess cannot even be evaluated. Yet the integral is a perfectly good (negative) area. The fix: for x < 0, use ln(−x), whose derivative is:

Intuition: 1/x is an odd function with two separate branches, and its antiderivative must be built branch by branch — ln|x| is exactly that two-branch construction, stitched into a single expression.

Derivation

We verify F(x) = ln|x| branch by branch. The absolute value is defined piecewise, so the verification splits the same way — and both branches land on 1/x.

A subtlety, honestly noted: because 1/x lives on two disconnected intervals, the “constant” C can technically be a different constant on each side. Calculus I courses write a single C, which is correct whenever you stay on one side of zero — and that covers every problem on the exam.

How to use it

Before reading on: if the answer really is ln|x| + C, differentiating must return 1/x on both sides of zero. Which differentiation rule will the x < 0 branch force you to use?

The procedure, every time:

1. Spot exactly 1/x (or x⁻¹). Not 1/x², not 1/(x+1) — exactly 1/x, possibly times a constant.
2. Write ln|x| + C. The bars are mandatory, not optional.
3. Constants factor out: ∫ (4/x) dx = 4 ln|x| + C. And ∫ dx/(3x) = (1/3) ln|x| + C — pull the 1/3 out first.
4. Simplify quotients before integrating: (x+1)/x = 1 + 1/x. Never integrate a fraction as a lump — split it into a sum first.
5. Check by differentiating: d/dx [ln|x|] should give back 1/x.

Reciprocal or power rule?

This is the #1 judgment call of the chapter. Only the exact exponent −1 gets the logarithm. 1/x² = x⁻² → power rule: −1/x + C. 1/√x = x^−1/2 → power rule: 2√x + C. Drill the pair: “negative one is log; everything else is power.”

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Memorization tips

• Say it aloud: “one over x integrates to log-abs-x.” Three beats, in order: reciprocal, log, bars.
• Drill the pair: “negative one is log; everything else is power.” The single most-tested distinction in the chapter.
• The bars test: if your antiderivative would break on negative x while the integrand survives there, you dropped the | |.
• Anchor on e: ∫₁^e dx/x = 1. One clean definite integral that ties the log to its base.
• Rewrite reflex: 1/(3x) = (1/3)·(1/x). Pull constants out before reaching for the rule.
• Quotient alarm: (x²+1)/x is not “one over something” — split it into x + 1/x first, then integrate each piece.

Final challenge

Five mixed questions — negative bounds, the power-rule trap, and splits. Score 5/5 and the reciprocal is yours.