Difficulty: Intermediate | Prerequisites: Derivatives of ln(x) and e^x, basic antiderivatives, u-substitution fundamentals, trig identities.
This set of problems covers the integration side of exponential and logarithmic calculus. The central technique is u-substitution, which reverses the chain rule: you spot an inside function, set u equal to it, convert the integral, and evaluate. The problems range from straightforward substitutions (1/(3x+4), powers of ln(x)) through polynomial long division before integrating, trig integrals that produce logarithmic results, and applied problems using the Fundamental Theorem of Calculus to recover a function from its derivative. You should already be comfortable with the derivative rules covered in the companion derivatives notes and with basic trig identities (tan = sin/cos, sec² = 1 + tan²).
Most of these integrals boil down to one move: pick the right u, find du, adjust constants, and integrate. When the numerator's degree is at least the denominator's, do polynomial long division first. For applied problems, the FTC lets you recover function values from a known derivative and an initial condition by computing a definite integral.
u-substitution
The integration counterpart of the chain rule. You identify an inside function u, compute du, rewrite the integral entirely in terms of u and du, integrate, then substitute back. Think of it as undoing the chain rule.
Polynomial long division (for integrals)
When the degree of the numerator is greater than or equal to the degree of the denominator, divide first to get a polynomial plus a simpler fraction. In simple terms, you simplify the fraction before integrating because you cannot directly integrate a top-heavy rational expression.
Definite integral with substitution
When using u-sub on a definite integral, convert the limits of integration to u-values. This avoids having to substitute back to x before evaluating.
Fundamental Theorem of Calculus (applied)
If you know f'(x) and one value f(a), you can find f(b) using f(b) = f(a) + integral from a to b of f'(x) dx. In simple terms, the integral of the rate of change gives you the total change.
Integral of 1/(3x + 4) dx: set u = 3x + 4, du = 3 dx, so dx = (1/3) du. The integral becomes (1/3) integral of du/u = (1/3) ln|3x + 4| + C.
Pattern: whenever you see 1/(linear function), set u = that linear function. The coefficient out front is always 1/(coefficient of x).
Integral from e to e² of (ln x)⁴ / x dx: set u = ln x, du = (1/x) dx. When x = e, u = 1. When x = e², u = 2. The integral becomes integral from 1 to 2 of u⁴ du = (1/5)u⁵ evaluated from 1 to 2 = 32/5 - 1/5 = 31/5.
Pattern: (ln x)^n / x is always a u-sub with u = ln x, because du = (1/x) dx is sitting right there.
Integral from 0 to 3 of (x² - 5)/(x + 2) dx: the numerator has higher degree, so divide first. x² - 5 divided by (x + 2) gives (x - 2) with remainder -1. So the integrand becomes (x - 2) - 1/(x + 2).
Integrate term by term: (1/2)x² - 2x - ln|x + 2|, evaluated from 0 to 3.
At x = 3: 9/2 - 6 - ln 5. At x = 0: 0 - 0 - ln 2. Result: 9/2 - 6 - ln 5 + ln 2 = -3/2 + ln(2/5).
Integral of tan(3x) dx: rewrite as sin(3x)/cos(3x) dx. Set u = cos(3x), du = -3 sin(3x) dx. The integral becomes -(1/3) ln|cos(3x)| + C, or equivalently (1/3) ln|sec(3x)| + C.
Integral of x² sec(x³) dx: set u = x³, du = 3x² dx. The integral becomes (1/3) integral of sec(u) du = (1/3) ln|sec(x³) + tan(x³)| + C.
Integral of x(x² + 4)³ dx: set u = x² + 4, du = 2x dx. The integral becomes (1/2) integral of u³ du = (1/2)(1/4)u⁴ + C = (1/8)(x² + 4)⁴ + C.
Integral of x² cos(2x³) dx: set u = 2x³, du = 6x² dx. The integral becomes (1/6) integral of cos(u) du = (1/6) sin(2x³) + C.
Integral from pi/8 to pi/6 of cos²(2x) sin(2x) dx: set u = cos(2x), du = -2 sin(2x) dx. Change limits: when x = pi/8, u = cos(pi/4) = sqrt(2)/2. When x = pi/6, u = cos(pi/3) = 1/2.
The integral becomes -(1/2) integral from sqrt(2)/2 to 1/2 of u² du = -(1/6)u³ evaluated from sqrt(2)/2 to 1/2.
Result: -(1/6)(1/8) + (1/6)(sqrt(2)/2)³ = -1/48 + (2 sqrt(2))/48 = (2 sqrt(2) - 1)/48.
Recovering f(x) from f'(x) and an initial value: if f'(x) = sin(x²) and f(1) = -3, find f(2). Use f(2) = f(1) + integral from 1 to 2 of sin(x²) dx. The integral evaluates numerically to approximately 0.4945, so f(2) is approximately -2.506.
Running the FTC backwards: if f'(x) = e^(-x²) and f(3) = -1, find f(2). Use integral from 2 to 3 of f'(x) dx = f(3) - f(2), so f(2) = f(3) - integral from 2 to 3 of e^(-x²) dx. The integral is approximately 0.00412, giving f(2) approximately -1.004.
Coffee starts at 93 degrees C in a 20 degree C room. The rate of temperature drop is r(t) = 5e^(-0.3t) degrees C per minute. To find the temperature at t = 8 minutes, compute 93 - integral from 0 to 8 of 5e^(-0.3t) dt.
The integral evaluates to approximately 15.155, giving a temperature of approximately 77.845 degrees C.
\int \frac{1}{u}\,du = \ln|u| + C\int \tan(u)\,du = -\ln|\cos(u)| + C = \ln|\sec(u)| + C\int \sec(u)\,du = \ln|\sec(u) + \tan(u)| + Cf(b) = f(a) + \int_a^b f'(x)\,dx\int_0^T r(t)\,dt = \text{total change over } [0, T]U-sub checklist: identify u, compute du, check that the remaining x-terms match du (up to a constant), convert, integrate, substitute back (or convert limits for definite integrals).
The integral of e^(-kt) models decay processes: how much of a drug has been absorbed, how much heat a cooling object has lost, how much charge remains on a discharging capacitor. The FTC technique of recovering a quantity from its rate of change is precisely how engineers compute total displacement from velocity data, total energy from power curves, and total cost from marginal cost functions.
Students often forget to adjust the constant when doing u-sub. If u = 3x + 4 and du = 3 dx, then dx = (1/3) du, not du. Missing the 1/3 factor is the most common error.
Trying to integrate (x² - 5)/(x + 2) directly without long division. If the numerator's degree equals or exceeds the denominator's, you must divide first.
Forgetting to change the limits of integration when doing u-sub on a definite integral. If you convert to u but keep the x-limits, the answer will be wrong.
Confusing the sign in the integral of tan(u). The result is -ln|cos(u)| + C, not +ln|cos(u)| + C. The negative comes from the -sin in the derivative of cos.
In applied FTC problems, subtracting the integral from the initial value when the rate gives a decrease (like cooling), but adding it when the problem says "dropping at a rate of": if r(t) is the rate of drop, then T(8) = T(0) - integral of r(t) dt.
⚠️ u-substitution is the single most tested integration technique at this level. Every exam will have several problems requiring it.
⚠️ Polynomial long division before integrating is a favourite exam trick. If you see a rational function where the top degree meets or exceeds the bottom, divide first.
⚠️ Changing limits in definite integrals with u-sub is a common source of lost marks. Practice converting limits until it is automatic.
⚠️ Applied FTC problems ("given f'(x) and f(a), find f(b)") appear on nearly every calculus final. Know the formula f(b) = f(a) + integral from a to b of f'(x) dx cold.
⚠️ The coffee-cooling problem is a classic Newton's Law of Cooling application. Expect at least one word problem that gives a rate function and asks for total change.
True or False: integral of 1/(5x + 3) dx = ln|5x + 3| + C. (False: it equals (1/5) ln|5x + 3| + C.)
Fill in the blank: for integral of (ln x)³ / x dx, set u = ______. (ln x)
True or False: when evaluating a definite integral by u-sub, you can keep the original x-limits if you substitute back to x before evaluating. (True.)
Fill in the blank: integral of tan(x) dx = ______ + C. (-ln|cos(x)| or ln|sec(x)|)
True or False: if f'(x) = e^(-x²) and f(3) = -1, then f(2) = -1 + integral from 2 to 3 of e^(-x²) dx. (False: f(2) = f(3) - integral from 2 to 3 of f'(x) dx = -1 - integral from 2 to 3 of e^(-x²) dx.)
Q: Evaluate integral of 1/(3x + 4) dx.
A: Set u = 3x + 4, du = 3 dx. Result: (1/3) ln|3x + 4| + C.
Q: Evaluate integral from e to e² of (ln x)⁴ / x dx.
A: u = ln x, du = (1/x) dx. Limits become 1 to 2. Integral of u⁴ du from 1 to 2 = (1/5)(32 - 1) = 31/5.
Q: Evaluate integral from 0 to 3 of (x² - 5)/(x + 2) dx.
A: Long division gives (x - 2) - 1/(x + 2). Integrate: (1/2)x² - 2x - ln|x + 2| from 0 to 3 = -3/2 + ln(2/5).
Q: Evaluate integral of tan(3x) dx.
A: Rewrite as sin(3x)/cos(3x), set u = cos(3x). Result: -(1/3) ln|cos(3x)| + C, or (1/3) ln|sec(3x)| + C.
Q: Evaluate integral of x(x² + 4)³ dx.
A: u = x² + 4, du = 2x dx. Result: (1/8)(x² + 4)⁴ + C.
Q: If f'(x) = sin(x²) and f(1) = -3, find f(2) to three decimal places.
A: f(2) = f(1) + integral from 1 to 2 of sin(x²) dx = -3 + 0.4945 = -2.506 (by calculator).
Q: A cup of coffee at 93 degrees C is cooling at a rate r(t) = 5e^(-0.3t) degrees C per minute in a 20 degree C room. Estimate the temperature at t = 8 minutes.
A: T(8) = 93 - integral from 0 to 8 of 5e^(-0.3t) dt = 93 - 15.155 = 77.845 degrees C (approximately).
U-substitution is the foundation for every integration technique that follows: integration by parts, trig substitution, and partial fractions all build on the ability to spot and execute a substitution cleanly. The applied FTC problems connect directly to differential equations, where you solve dy/dx = f(x) by integrating both sides and using an initial condition. The cooling problem is a gateway to Newton's Law of Cooling as a separable differential equation, which you will likely see in a later chapter.
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