The arctan form

Where it comes from

The problem is the reverse of d/dx [arctan x] = 1/(1+x²) — the inverse-trig derivative. It is the first integral on the sheet whose answer is not built from powers, exponentials, logs, or plain trig: the antiderivative is an inverse trig function.

The tempting log guess:

Before reading on: the numerator here is 1, but d/dx [1+x²] = 2x. The log rule needs the numerator to be the derivative of the denominator. Is it? What does that tell you about the ln guess?

It looks like the f′/f pattern (∫ f′/f = ln|f|). But that pattern demands the numerator be the derivative of the denominator — and d/dx [1+x²] = 2x, not 1. Kill it by differentiating:

Intuition: 1/(1+x²) is a bell-shaped curve (1 at x = 0, decaying to 0 both ways). Its total area is π — and indeed ∫_−∞^∞ dx/(1+x²) = π. An antiderivative must flatten out at both ends, approaching finite limits — exactly how arctan x behaves, leveling off at ±π/2. The geometry already whispers “inverse tangent.”

Derivation

We prove d/dx [arctan x] = 1/(1+x²) by implicit differentiation — the standard way to differentiate inverse functions. Then the integral is that fact read backwards.

Why +C is the whole story: any two antiderivatives differ by a constant (zero derivative ⇒ constant), so arctan x + C captures every one of them.

How to use it

Before reading on: ∫ dx/(1+4x²) hides a 2x inside the denominator. Before the reveal: will the answer be exactly arctan(2x) + C, or does the hidden 2x force an extra factor — and which way does it go?

The procedure, every time:

1. Spot the pattern: a constant numerator over 1 + (something)². The denominator must be a sum — 1 + x², not 1 − x², not √(1+x²).
2. Plain 1 + x²: write arctan x + C.
3. Scaled form 1/(a² + x²): ∫ dx/(a²+x²) = (1/a)·arctan(x/a) + C. Check by differentiating — the chain rule supplies exactly the 1/a factors.
4. Constants factor out: ∫ 2/(1+x²) dx = 2 arctan x + C.

Arctan or something else?

Numerator 2x: ∫ 2x/(1+x²) dx = ln(1+x²) + C — now the log pattern fires, because 2x is the derivative of 1+x². Minus sign: 1/(1−x²) is not arctan territory (it needs partial fractions). Square root: 1/√(1−x²) gives arcsin x, the sibling inverse-trig integral. Drill the trio: plus-and-no-root → arctan; minus-under-root → arcsin; numerator-is-derivative → log.

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 one-plus-x-squared is arctan.” The denominator is the trigger phrase.
• Drill the trio: 1 upstairs → arctan; derivative upstairs → log; minus under a root → arcsin. One glance at the numerator sorts them.
• Anchor: ∫₀¹ dx/(1+x²) = π/4. The most famous definite integral on the sheet.
• The scaled form: ∫ dx/(a²+x²) = (1/a) arctan(x/a) + C — “one over a, arctan of x over a.”
• The numerator test: before writing arctan, confirm the numerator is constant. If it contains x, stop and re-read — it might be the log pattern instead.

Final challenge

Five mixed questions — the log trap, symmetric bounds, and the scaled form. Score 5/5 and the arctan form is yours.