Lesson 31: Integrating Composite, Exponential, Log & Inverse-Trig Forms

(a) Opening Question

You already know ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C — for every n except −1. So what is ∫ (1/x) dx?

The power rule breaks (it would put x⁰/0, undefined). You need a different reflex. Ask the question that drives this entire lesson:

What function has 1/x as its derivative?

You learned in Unit 3 that d/dx[ln x] = 1/x. So ∫ (1/x) dx = ln|x| + C. (The absolute value lets the formula work for negative x too — we'll see why it matters.)

Now stretch it. Evaluate:

∫ (2x)/(x² + 1) dx

Notice the numerator 2x is exactly the derivative of the denominator x² + 1. That's not a coincidence you set up with u-substitution — it's a pattern to recognize. The answer is ln(x² + 1) + C. This lesson is about training that recognition across logs, exponentials, and inverse-trig forms.

(b) Core Concepts

Every antiderivative in this lesson is a derivative rule read backwards. You are not memorizing new integrals so much as recognizing old derivatives. The single most useful habit on the AP exam: before reaching for u-substitution, ask "what function, when differentiated, produces this integrand?"

The form table

Memorize these. On the no-calculator section there is no other way to get them.

Logarithm form

∫ (1/x) dx = ln|x| + C
∫ g'(x)/g(x) dx = ln|g(x)| + C

Exponential forms

∫ eˣ dx = eˣ + C
∫ eᵍ⁽ˣ⁾ g'(x) dx = eᵍ⁽ˣ⁾ + C
∫ aˣ dx = aˣ / ln a + C        (a > 0, a ≠ 1)

Inverse-trig forms (yes, these are on AB)

∫ dx/(1 + x²) = arctan x + C
∫ dx/√(1 − x²) = arcsin x + C
∫ dx/(a² + x²) = (1/a) arctan(x/a) + C
∫ dx/√(a² − x²) = arcsin(x/a) + C

There is also ∫ dx/(x√(x²−1)) = arcsec|x| + C, but it is rare on AB — focus your energy on arctan and arcsin.

Why ln|g|, with the absolute value

The derivative d/dx[ln|x|] = 1/x holds for all x ≠ 0, not just positive x. The integrand g'/g can be negative, so the antiderivative must be defined there too. Dropping the absolute value is a scoring trap — write ln|g(x)| every time unless you can prove g(x) > 0 on the interval.

A worked log form. Find ∫ tan x dx. Rewrite tan x = sin x / cos x. The numerator sin x is almost the derivative of the denominator cos x — off by a sign, since d/dx[cos x] = −sin x. So:

∫ tan x dx = ∫ (sin x)/(cos x) dx = −∫ (−sin x)/(cos x) dx = −ln|cos x| + C

Check by differentiating: d/dx[−ln|cos x|] = −(1/cos x)(−sin x) = sin x/cos x = tan x. ✓ This is a standard result worth memorizing: ∫ tan x dx = −ln|cos x| + C.

A worked inverse-trig form (matching a²)

Find ∫ dx/(x² + 9). This matches ∫ dx/(a² + x²) with a² = 9, so a = 3:

∫ dx/(x² + 9) = (1/3) arctan(x/3) + C

Check: d/dx[(1/3)arctan(x/3)] = (1/3) · [1/(1 + (x/3)²)] · (1/3) = (1/9)/(1 + x²/9) = 1/(9 + x²). ✓ The two factors of 1/3 are the trap — students who write arctan(x/3) without the leading 1/3 lose the point.

Rewriting before integrating

Some integrands don't look like any form until you do algebra first.

Long division (improper rational integrand — degree of top ≥ degree of bottom). Find ∫ x²/(x² + 1) dx. Since the degrees are equal, divide first:

x²/(x² + 1) = (x² + 1 − 1)/(x² + 1) = 1 − 1/(x² + 1)

Now each piece is standard:

∫ x²/(x² + 1) dx = ∫ 1 dx − ∫ dx/(x² + 1) = x − arctan x + C

Never try to integrate an improper rational expression directly — divide first. Quick test: if the top's degree is ≥ the bottom's, long-divide.

Completing the square (to reach arctan). Find ∫ dx/(x² + 4x + 13). The denominator doesn't factor nicely over the reals; complete the square:

x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 9

Now it matches the arctan form with a = 3 and the shift u = x + 2:

∫ dx/((x + 2)² + 9) = (1/3) arctan((x + 2)/3) + C

Check: differentiating (1/3)arctan((x+2)/3) gives (1/3)·[1/(1+((x+2)/3)²)]·(1/3) = 1/((x+2)²+9) = 1/(x²+4x+13). ✓

Splitting fractions

When a single fraction is really a sum, split it and integrate term by term. For ∫ (x + 1)/(x² + 1) dx, split into two recognizable forms:

(x + 1)/(x² + 1) = x/(x² + 1) + 1/(x² + 1)

The first is a log form (numerator is half the derivative of the bottom): ∫ x/(x²+1) dx = (1/2)ln(x²+1). The second is the arctan form: ∫ dx/(x²+1) = arctan x. So the whole integral is (1/2)ln(x² + 1) + arctan x + C. One term log, one term inverse-trig — splitting is what reveals both.

The decision habit

When you see a rational or composite integrand, run this checklist:

1. Is the numerator the derivative of the denominator? → log form, ln|g| + C.
2. Is there an inner function with its derivative present? → eᵍ, or u-sub.
3. Is it 1/(a² + x²) or 1/√(a² − x²)? → arctan or arcsin.
4. Is the rational integrand improper (top degree ≥ bottom)? → long-divide first.
5. Quadratic in the denominator that won't factor? → complete the square, expect arctan.

(c) Worked Examples

Example 1 — Log form g'/g (NO CALC, foundation)

Problem. Evaluate ∫ x/(x² + 4) dx.

Strategy. Check the log pattern: is the numerator a constant multiple of the derivative of the denominator? d/dx[x² + 4] = 2x, and the numerator is x = (1/2)(2x). Yes — pull out the 1/2.

Solution.

∫ x/(x² + 4) dx = (1/2) ∫ (2x)/(x² + 4) dx = (1/2) ln|x² + 4| + C = (1/2) ln(x² + 4) + C

(Here x² + 4 > 0 always, so the absolute value is optional — but writing it is never wrong.)

Verify. d/dx[(1/2)ln(x²+4)] = (1/2)·(2x)/(x²+4) = x/(x²+4). ✓

Example 2 — Exponential composite (NO CALC, procedural)

Problem. Evaluate ∫ x e^(x²) dx.

Strategy. The inner function is g = x² with g' = 2x. The integrand has an x, which is (1/2)g'. This is the ∫ eᵍ g' dx form once you supply the missing constant.

Solution.

∫ x e^(x²) dx = (1/2) ∫ e^(x²) · 2x dx = (1/2) e^(x²) + C

Verify. d/dx[(1/2)e^(x²)] = (1/2)e^(x²)·2x = x e^(x²). ✓

Example 3 — Inverse-trig form with a² (NO CALC, AP level)

Problem. Evaluate ∫₀³ dx/(x² + 9).

Strategy. Matches ∫ dx/(a² + x²) with a = 3. Find the antiderivative, then apply FTC.

Solution.

∫ dx/(x² + 9) = (1/3) arctan(x/3) + C

∫₀³ dx/(x² + 9) = [ (1/3) arctan(x/3) ]₀³
              = (1/3) arctan(1) − (1/3) arctan(0)
              = (1/3)(π/4) − 0
              = π/12

Justification of the bound values. arctan(3/3) = arctan(1) = π/4 and arctan(0) = 0. Answer: π/12.

Verify (antiderivative): d/dx[(1/3)arctan(x/3)] = (1/3)·(1/3)/(1 + x²/9) = 1/(9 + x²). ✓

Example 4 — Long division + completing the square (NO CALC, AP level)

Problem. Evaluate ∫ (x² + 4x + 6)/(x² + 4x + 13) dx.

Strategy. Top and bottom have equal degree → the integrand is improper. Long-divide. The remainder will land on a complete-the-square arctan form.

Solution. Since the leading terms match, subtract the denominator from the numerator:

(x² + 4x + 6)/(x² + 4x + 13) = (x² + 4x + 13 − 7)/(x² + 4x + 13) = 1 − 7/(x² + 4x + 13)

Complete the square in the leftover denominator: x² + 4x + 13 = (x + 2)² + 9. Then:

∫ (x² + 4x + 6)/(x² + 4x + 13) dx = ∫ 1 dx − 7 ∫ dx/((x + 2)² + 9)
                                  = x − 7 · (1/3) arctan((x + 2)/3) + C
                                  = x − (7/3) arctan((x + 2)/3) + C

Verify. d/dx[x − (7/3)arctan((x+2)/3)] = 1 − (7/3)·(1/3)/(1 + ((x+2)/3)²) = 1 − 7/((x+2)²+9) = 1 − 7/(x²+4x+13), which equals (x²+4x+6)/(x²+4x+13). ✓

(d) Common Mistakes

Dropping the absolute value in ln|g|. Students write ∫ tan x dx = −ln(cos x) + C. Why it's wrong: cos x is negative on parts of the domain, where ln(cos x) is undefined. Fix: always write ln|g(x)| unless you can guarantee g > 0 on the interval (e.g. x² + 1).

Confusing the arcsin and arctan forms. Students see a denominator and guess. Why it's wrong: 1/(a² + x²) → arctan, but 1/√(a² − x²) → arcsin. The square root and the minus sign are the tells. Fix: memorize "square root + subtraction = arcsin; plain sum = arctan." A bare a² − x² with no root is neither — it's a partial-fractions problem (not AB).

Integrating an improper rational expression directly. Students try to force ∫ x²/(x² + 1) dx straight into a form. Why it's wrong: there's no antiderivative form for it as written. Fix: if the numerator's degree ≥ the denominator's, long-divide first. Here it becomes x − arctan x + C.

Dropping the 1/a factor. Students write ∫ dx/(x² + 9) = arctan(x/3) + C, forgetting the leading 1/3. Why it's wrong: the chain rule from arctan(x/3) produces an extra 1/3 that must be cancelled. Fix: the answer is (1/3)arctan(x/3) + C. Always differentiate to confirm the constant.

Forgetting 1/ln a for ∫ aˣ dx. Students write ∫ 2ˣ dx = 2ˣ + C. Why it's wrong: d/dx[2ˣ] = 2ˣ ln 2, so the antiderivative needs the 1/ln 2 to cancel it. Fix: ∫ aˣ dx = aˣ/ln a + C.

(e) Practice Problems

Mostly non-calculator. Find a general antiderivative (with + C) unless bounds are given.

∫ (cos x)/(2 + sin x) dx =

  (A) ln|2 + sin x| + C  (B) −ln|2 + sin x| + C  (C) (sin x)/(2 + sin x) + C  (D) arctan(sin x) + C

∫ e^(3x) dx =

  (A) e^(3x) + C  (B) 3e^(3x) + C  (C) (1/3)e^(3x) + C  (D) e^(3x)/(3x) + C

∫₀¹ dx/(1 + x²) =

  (A) π/2  (B) π/4  (C) 1  (D) ln 2

∫ dx/(25 + x²) =

  (A) arctan(x/5) + C  (B) (1/5)arctan(x/5) + C  (C) 5 arctan(5x) + C  (D) (1/5)arcsin(x/5) + C

∫ dx/√(9 − x²) =

  (A) arctan(x/3) + C  (B) (1/3)arcsin(x/3) + C  (C) arcsin(x/3) + C  (D) ln|9 − x²| + C

∫ 4ˣ dx =

  (A) 4ˣ + C  (B) 4ˣ ln 4 + C  (C) 4ˣ/ln 4 + C  (D) x·4ˣ⁻¹ + C

∫ x²/(x² + 4) dx =

  (A) x − 2 arctan(x/2) + C  (B) x − (1/2)arctan(x/2) + C  (C) (1/2)arctan(x/2) + C  (D) ln(x² + 4) + C

∫ dx/(x² − 6x + 13) =

  (A) (1/2)arctan((x − 3)/2) + C  (B) arctan(x − 3) + C  (C) (1/2)ln|x² − 6x + 13| + C  (D) arcsin((x − 3)/2) + C

∫ (3x²)/(x³ + 1) dx =

  (A) ln|x³ + 1| + C  (B) (1/3)ln|x³ + 1| + C  (C) 3 ln|x³ + 1| + C  (D) arctan(x³) + C

Short answer — show all work; no calculator.

Evaluate ∫ (sec² x)/(1 + tan x) dx.

Evaluate ∫ x e^(x²) dx and evaluate ∫₀¹ 2x/(x² + 1) dx.

Evaluate ∫ (x + 1)/(x² + 1) dx. (Hint: split the fraction.)

[Justify] A student claims ∫ dx/(4 + x²) = (1/2)arcsin(x/2) + C. Identify the error and give the correct antiderivative, justifying which form applies and why.

[Justify] Evaluate ∫ e^x/(1 + e^x) dx. State explicitly why this is a log form (identify g and g'), then justify whether the absolute value on the logarithm is required here.

(f) AP Exam Focus — Free Response

> No calculator. Show all work. Answers must be exact (in terms of π and ln). (9 points total)

Water flows into a reservoir at a rate modeled by

R(t) = 2t/(t² + 1) + 6/(t² + 9)

thousand gallons per hour, where t is measured in hours for 0 ≤ t ≤ 3. At time t = 0 the reservoir contains 5 thousand gallons of water.

(a) (3 points) Find the total amount of water that flows into the reservoir during the time interval 0 ≤ t ≤ 3. Show the setup and give an exact answer.

(b) (2 points) Write, but do not evaluate, an expression for the total amount of water W(t) in the reservoir at time t. Then find W(3), the exact amount of water in the reservoir at t = 3.

(c) (2 points) Find the average rate at which water flows into the reservoir over 0 ≤ t ≤ 3. Indicate units of measure.

(d) (2 points) Is the rate of inflow increasing or decreasing at t = 1? Justify your answer.

Model Solution

(a) The total water in is the integral of the rate:

∫₀³ R(t) dt = ∫₀³ [ 2t/(t² + 1) + 6/(t² + 9) ] dt

Integrate term by term. The first is a log form (g = t² + 1, g' = 2t); the second is an arctan form (a = 3):

∫₀³ 2t/(t² + 1) dt = [ ln(t² + 1) ]₀³ = ln 10 − ln 1 = ln 10

∫₀³ 6/(t² + 9) dt = 6 · (1/3)[ arctan(t/3) ]₀³ = 2[ arctan 1 − arctan 0 ] = 2 · π/4 = π/2

Total = ln 10 + π/2 thousand gallons. (≈ 3.873 thousand gallons.)

(b) Starting amount plus accumulated inflow:

W(t) = 5 + ∫₀ᵗ R(x) dx        (do not evaluate)

Then, using part (a),

W(3) = 5 + (ln 10 + π/2) = 5 + ln 10 + π/2 thousand gallons.

(c) The average rate of inflow over [0, 3] is the average value of R:

average rate = (1/(3 − 0)) ∫₀³ R(t) dt = (1/3)(ln 10 + π/2)
            = (ln 10)/3 + π/6 thousand gallons per hour.

Units: thousand gallons per hour. (≈ 1.291 thousand gallons per hour.)

(d) Differentiate R and evaluate the sign at t = 1.

R(t) = 2t(t² + 1)⁻¹ + 6(t² + 9)⁻¹

R'(t) = 2·(t²+1)⁻¹ + 2t·(−1)(t²+1)⁻²(2t) + 6·(−1)(t²+9)⁻²(2t)
      = 2/(t²+1) − 4t²/(t²+1)² − 12t/(t²+9)²

At t = 1: t² + 1 = 2 and t² + 9 = 10, so

R'(1) = 2/2 − 4/4 − 12/100 = 1 − 1 − 0.12 = −0.12 < 0.

Since R'(1) < 0, the rate of inflow is decreasing at t = 1.

Scoring Commentary

• Part (a): 1 pt for the integral setup ∫₀³ R(t) dt; 1 pt for the log antiderivative ln(t²+1); 1 pt for the arctan antiderivative 2 arctan(t/3) with the 1/3 factor handled and the exact value ln 10 + π/2. Students who write arctan(t/3) without the leading (6)(1/3) = 2 lose the final point — the most common error here.
• Part (b): 1 pt for the integral expression with the initial condition 5 +; 1 pt for W(3) = 5 + ln 10 + π/2. A "do not evaluate" part still requires correct bounds and the +5.
• Part (c): 1 pt for the average-value setup (1/3)∫₀³ R dt; 1 pt for the exact value with units. Omitting units is a frequent lost point.
• Part (d): 1 pt for evaluating R'(1) (or otherwise determining the sign of R' at t=1); 1 pt for the conclusion "decreasing" tied explicitly to R'(1) < 0. Stating "decreasing" without the supporting sign of R' earns 0 for the justification — AP requires the reason, not just the answer.

∫ (sec² x)/(1 + tan x) dx = ln|1 + tan x| + C

∫ x e^(x²) dx = (1/2)e^(x²) + C

∫₀¹ 2x/(x² + 1) dx = [ ln(x² + 1) ]₀¹ = ln 2 − ln 1 = ln 2

(x + 1)/(x² + 1) = x/(x² + 1) + 1/(x² + 1)

∫ (x + 1)/(x² + 1) dx = (1/2)ln(x² + 1) + arctan x + C

∫ dx/(4 + x²) = (1/2)arctan(x/2) + C

∫ e^x/(1 + e^x) dx = ln|1 + e^x| + C = ln(1 + e^x) + C