This set covers the core toolkit for evaluating integrals that resist basic antidifferentiation: u-substitution (including arctan and exponential forms), polynomial long division, completing the square, trigonometric substitution, and partial-fraction decomposition. These are the techniques you reach for when the integrand does not match a standard form directly, and they appear heavily on exams.
Difficulty: Intermediate. Prerequisites: comfort with basic antiderivatives, the chain rule in reverse, and standard trig identities (Pythagorean identity, double-angle formulas). You should also be familiar with the definitions of arctan and arcsin, and with the natural logarithm, before tackling this material.
Integral calculus, at its heart, is about undoing differentiation. The trouble is that most functions you meet in practice do not look like anything you differentiated before. Integration techniques are the algebraic and trigonometric manoeuvres that reshape an integrand into a form whose antiderivative you recognise.
This material sits squarely in the middle of a first integral calculus course and feeds directly into later topics: improper integrals, applications of integration (area, volume, arc length), and series convergence tests all assume you can evaluate integrals fluently. If you are behind, make sure you are comfortable with derivatives of inverse trig functions and with logarithmic differentiation before proceeding.
u-substitution
The integration analogue of the chain rule. You identify an inner function u = g(x) whose derivative g'(x) appears (up to a constant) elsewhere in the integrand, rewrite everything in terms of u and du, integrate, then substitute back. Think of it as: you are choosing a new variable that makes the integral look like a standard form.
Polynomial long division
An algebraic step applied before integration when the degree of the numerator is greater than or equal to the degree of the denominator in a rational function. You divide to express the integrand as a polynomial plus a proper fraction, then integrate each piece separately. In simple terms, this means you are splitting a "top-heavy" fraction into parts you already know how to integrate.
Completing the square
A technique for rewriting a quadratic expression ax^2 + bx + c in the form a(x + h)^2 + k. This is essential when the integrand contains a square root of a quadratic or a quadratic in the denominator, because the completed form matches the arcsin or arctan standard integrals. Think of it as reshaping the quadratic so you can see which trig substitution or inverse-trig formula applies.
Trigonometric substitution
A method for integrals involving sqrt(a^2 - x^2), sqrt(a^2 + x^2), or sqrt(x^2 - a^2). You replace x with a trig expression (a sin u, a tan u, or a sec u) so the square root simplifies via a Pythagorean identity. In simple terms, you trade an algebraic mess for a trig integral that is often much cleaner.
Partial-fraction decomposition
A technique for integrating rational functions by breaking them into a sum of simpler fractions whose denominators are linear or irreducible quadratic factors. Each simpler fraction integrates to a logarithm, an arctangent, or a power. Think of it as: you are un-adding fractions so each piece is easy on its own.
Rationalising the integrand
Multiplying numerator and denominator by a conjugate or equivalent expression to eliminate a radical from a fraction before integrating. This converts the integrand into a form where substitution or standard formulas apply directly.
The central idea is pattern recognition: spot an inner function whose derivative already sits in the integrand.
Inverse-trig substitution (arctan form)
When the integrand contains 1/(1 + y^2), look for u = arctan(y), so du = 1/(1 + y^2) dy.
The integral of e^(3 arctan y) / (1 + y^2) dy becomes the integral of e^(3u) du after substitution.
The antiderivative of e^(3u) is e^(3u)/3. Evaluate at the new limits: arctan(1) = pi/4, arctan(-1) = -pi/4.
Result: (1/3)(e^(3pi/4) - e^(-3pi/4)).
Derivative-sitting-in-the-integrand pattern
For the integral of (x + 3) / sqrt(2 + 6x + x^2) dx, notice that d/dx(2 + 6x + x^2) = 2x + 6 = 2(x + 3).
Set u = 2 + 6x + x^2, then du = 2(x + 3) dx, so (x + 3) dx = du/2.
The integral becomes (1/2) times the integral of u^(-1/2) du = sqrt(u) + C = sqrt(2 + 6x + x^2) + C.
Exponential substitution
For integrals like (1 + 2e^x) / (1 - e^x) dx, set u = e^x so x = ln(u) and dx = du/u.
The integrand transforms into a rational function of u, which you then decompose using partial fractions.
Result: x - 3 ln|e^x - 1| + C.
Algebraic substitution (square root removal)
For the integral of sqrt(x - 1) / (2x) dx, set u^2 = x - 1 so 2u du = dx and x = u^2 + 1.
This converts the integral into u^2 / (u^2 + 1) du, a rational function you can split as 1 - 1/(u^2 + 1).
Integrating gives u - arctan(u), and back-substituting yields the answer.
When the numerator's degree equals or exceeds the denominator's degree, divide first. The quotient is a polynomial you integrate term by term; the remainder gives a proper fraction.
Worked example: integral of (x^2 + 4)/(x - 1) dx from 2 to 3
Rewrite the numerator: x^2 + 4 = (x^2 - 1) + 5 = (x - 1)(x + 1) + 5.
So (x^2 + 4)/(x - 1) = x + 1 + 5/(x - 1).
Integrate each piece: the integral of (x + 1) dx = x^2/2 + x, and the integral of 5/(x - 1) dx = 5 ln|x - 1|.
Evaluate from 2 to 3: (9/2 + 3 + 5 ln 2) - (2 + 2 + 0) = 7/2 + 5 ln 2.
When to use this
Check degrees first. If deg(numerator) >= deg(denominator), division is mandatory before any other technique.
After division, the remainder term may require further techniques (partial fractions, completing the square) depending on its form.
Completing the square converts a quadratic under a radical or in a denominator into a shifted-square form that matches a standard integral (arcsin, arctan, or a trig-sub template).
Worked example: integral of 5 / sqrt(5 - 4x - x^2) dx
Rewrite the quadratic: 5 - 4x - x^2 = -(x^2 + 4x - 5) = -(x^2 + 4x + 4 - 9) = 9 - (x + 2)^2.
The integral becomes 5 / sqrt(9 - (x + 2)^2) dx, which matches the arcsin template: integral of 1/sqrt(a^2 - w^2) dw = arcsin(w/a) + C.
Set x + 2 = 3 sin(u), so dx = 3 cos(u) du and sqrt(9 - (x + 2)^2) = 3 cos(u).
The integral simplifies to the integral of 5 du = 5u + C = 5 arcsin((x + 2)/3) + C.
How to recognise when you need it
The integrand has a quadratic ax^2 + bx + c under a square root, or in the denominator.
The quadratic is not already a perfect square or a simple a^2 - x^2 form.
Complete the square to reveal which standard form applies: if the completed form is a^2 - w^2, use arcsin or sin-substitution; if it is a^2 + w^2, use arctan.
Trig substitution eliminates square roots by exploiting Pythagorean identities. The three standard cases are outlined below.
sqrt(a^2 - x^2): use x = a sin(u)
Then sqrt(a^2 - x^2) = a cos(u) and dx = a cos(u) du.
Example: integral of (5 - x)/(1 + x^2) dx from 0 to 1. Set x = tan(u), so dx = sec^2(u) du and 1 + x^2 = sec^2(u). The integral splits into 5 times the integral of du minus the integral of tan(u) du, giving 5pi/4 - ln(sqrt(2)).
sqrt(a^2 + x^2): use x = a tan(u)
Then sqrt(a^2 + x^2) = a sec(u) and dx = a sec^2(u) du.
Used when the integrand has 1 + x^2 in the denominator without a square root: the tan substitution converts 1/(1 + x^2) into 1/sec^2(u), which cancels with the sec^2(u) from dx.
Rationalising before substitution
For sqrt((2 + x)/(2 - x)) dx from 0 to sqrt(3), rationalise the fraction: multiply top and bottom by sqrt(2 + x) to get (2 + x)/sqrt(4 - x^2).
Then use x = 2 sin(u), giving sqrt(4 - x^2) = 2 cos(u) and dx = 2 cos(u) du.
The integral becomes 2 times the integral of (1 + sin(u)) du from 0 to pi/3.
Result: 2(u - cos(u)) evaluated from 0 to pi/3 = 2pi/3 + 1.
Updating limits of integration
Whenever you change variables in a definite integral, convert the limits at the same time. This avoids having to substitute back into the original variable at the end.
Partial fractions apply when you have a proper rational function (degree of numerator less than degree of denominator) with factorable denominator.
Splitting into simpler fractions
For (1 + 2u)/((1 - u)u), rewrite the numerator as (1 - u) + 3u to get 1/u + 3/(1 - u).
Each piece integrates directly: ln|u| - 3 ln|1 - u| + C.
This technique arose in the exponential substitution example: after setting u = e^x, the integrand (1 + 2e^x)/(1 - e^x) became a rational function of u.
Algebraic rewriting before integration
For u^2/(u^2 + 1), write it as 1 - 1/(u^2 + 1). This avoids partial fractions entirely and gives a straightforward integral of 1 - arctan(u).
This pattern appears whenever the numerator's degree matches the denominator's: subtract and add a constant to split the fraction into a polynomial part and a proper part.
Splitting a fraction by separating terms in the numerator
For (5 - x)/(1 + x^2), write as 5/(1 + x^2) - x/(1 + x^2).
The first piece integrates to 5 arctan(x). The second integrates to (1/2) ln(1 + x^2) via u-substitution with u = 1 + x^2.
This additive splitting works whenever the denominator does not factor and the numerator has two terms with different character (a constant and a linear term over a quadratic).
\int \frac{1}{1+x^2}\,dx = \arctan(x) + C\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\!\left(\frac{x}{a}\right) + C\int \frac{x}{1+x^2}\,dx = \frac{1}{2}\ln(1+x^2) + C\int e^{au}\,du = \frac{e^{au}}{a} + C\int \tan(u)\,du = \ln|\sec(u)| + CPythagorean identities used in trig substitution
1 - sin^2(u) = cos^2(u)
1 + tan^2(u) = sec^2(u)
sec^2(u) - 1 = tan^2(u)
Completing the square template
ax^2 + bx + c = a(x + b/(2a))^2 + (c - b^2/(4a))
Students often forget to convert the limits of integration when switching variables in a definite integral. If you substitute u = arctan(y) and the original limits are y = -1 to y = 1, the new limits are u = -pi/4 to u = pi/4, not -1 to 1.
Students frequently skip polynomial long division and try to force partial fractions on an improper rational function. Partial fractions only work on proper fractions (degree of numerator strictly less than degree of denominator). Divide first.
When completing the square, students sometimes drop the sign. For 5 - 4x - x^2, the correct rewriting is 9 - (x + 2)^2, not 9 + (x + 2)^2. Getting the sign wrong leads to arctan when you should have arcsin, or vice versa.
Students confuse when to use sin-substitution versus tan-substitution. The rule: sqrt(a^2 - x^2) calls for x = a sin(u); a denominator of a^2 + x^2 (no square root) calls for x = a tan(u).
These techniques appear in nearly every integral calculus exam. Expect at least one problem requiring completing the square and one requiring polynomial division or partial fractions.
Trig substitution problems are a favourite for showing whether students understand the Pythagorean identity step. You may be asked to evaluate the integral and choose from multiple-choice answers that differ only by signs and coefficients.
The combination of u-substitution with limit conversion is heavily tested. Exams will penalise an answer left in terms of u when the question asked for evaluation in terms of x.
Know the three trig-sub cases cold: which substitution to use for each radical form. This decision step is where most marks are lost.
True or False: To integrate (x^3 + 2)/(x - 1) dx, you should use partial fractions directly. (Answer: False. The degree of the numerator exceeds that of the denominator, so you must divide first.)
Fill in the blank: When the integrand contains sqrt(9 - x^2), the appropriate substitution is x = ____. (Answer: 3 sin(u).)
True or False: Completing the square on 5 - 4x - x^2 gives (x - 2)^2 + 1. (Answer: False. It gives 9 - (x + 2)^2.)
Fill in the blank: If u = arctan(y), then du = ____. (Answer: dy/(1 + y^2).)
True or False: When you change variables in a definite integral, you must substitute the original limits back at the end. (Answer: False. You convert the limits to the new variable and evaluate directly.)
Q: Evaluate the integral of e^(3 arctan y) / (1 + y^2) dy from -1 to 1.
A: Substitute u = arctan(y), so du = dy/(1 + y^2). Limits become -pi/4 to pi/4. The integral becomes the integral of e^(3u) du = e^(3u)/3 evaluated at the limits, giving (1/3)(e^(3pi/4) - e^(-3pi/4)).
Q: Evaluate the integral of (x^2 + 4)/(x - 1) dx from 2 to 3.
A: Divide: (x^2 + 4)/(x - 1) = x + 1 + 5/(x - 1). Integrate: x^2/2 + x + 5 ln|x - 1|. Evaluate from 2 to 3 to get 7/2 + 5 ln 2.
Q: Determine the integral of 5/sqrt(5 - 4x - x^2) dx.
A: Complete the square: 5 - 4x - x^2 = 9 - (x + 2)^2. The integral becomes 5/sqrt(9 - (x + 2)^2) dx = 5 arcsin((x + 2)/3) + C.
Q: Evaluate the integral of (5 - x)/(1 + x^2) dx from 0 to 1.
A: Split the fraction: 5/(1 + x^2) - x/(1 + x^2). Integrate: 5 arctan(x) - (1/2) ln(1 + x^2). Evaluate from 0 to 1 to get 5pi/4 - ln(sqrt(2)).
Q: Evaluate the integral of sqrt((2 + x)/(2 - x)) dx from 0 to sqrt(3).
A: Rationalise to get (2 + x)/sqrt(4 - x^2). Substitute x = 2 sin(u). The integral becomes 2 times the integral of (1 + sin(u)) du from 0 to pi/3, giving 2pi/3 + 1.
These integration techniques feed directly into improper integrals, where you evaluate a definite integral with an infinite limit or a discontinuity, since the actual antidifferentiation step uses the same methods covered here.
Trig substitution reappears in applications of integration: computing arc length, surface area, and volumes of revolution often produces integrands with square roots of quadratics.
Partial fractions are also central to solving differential equations (separable and linear first-order), where you integrate both sides of a separated equation.
u-substitution, change of variables, integration by substitution, polynomial long division for integrals, completing the square integration, trig substitution, trigonometric substitution, inverse trig integrals, arctan integral, arcsin integral, partial fraction decomposition, rational function integration, rationalising the integrand, integral calculus, antiderivative techniques, UT Austin integral calculus, HW 07, Wolesensky