U-substitution, integration by parts (LIATE & tabular), trig integrals, trig substitution, partial fractions, and improper integrals — every method you need for Calculus 2.
Always check whether the integral matches a basic form before applying a technique. These are the antiderivatives every Calculus 2 student should know cold.
| Integral | Antiderivative |
|---|---|
| ∫xⁿ dx | xⁿ⁺¹/(n+1) + C (n ≠ −1) |
| ∫1/x dx | ln|x| + C |
| ∫eˣ dx | eˣ + C |
| ∫aˣ dx | aˣ / ln a + C |
| ∫sin x dx | −cos x + C |
| ∫cos x dx | sin x + C |
| ∫sec²x dx | tan x + C |
| ∫csc²x dx | −cot x + C |
| ∫sec x tan x dx | sec x + C |
| ∫csc x cot x dx | −csc x + C |
| ∫tan x dx | ln|sec x| + C |
| ∫sec x dx | ln|sec x + tan x| + C |
| ∫1/√(1−x²) dx | arcsin x + C |
| ∫1/(1+x²) dx | arctan x + C |
| ∫1/(x√(x²−1)) dx | arcsec|x| + C |
U-substitution is the chain rule in reverse. It is the first technique to try on any composite integral.
Identify u
Choose u = g(x) to be the inner function of a composition. Look for a function whose derivative also appears (or is a constant multiple of something) in the integrand.
Compute du
Differentiate: du = g'(x) dx. Solve for dx = du / g'(x) and substitute. If a constant factor is missing, adjust by multiplying and dividing.
Rewrite entirely in u
Every x and dx must be replaced. If any x remains, u was not chosen well — try a different substitution.
Integrate in u, back-substitute
Evaluate ∫f(u) du using the basic antiderivative table, then replace u with g(x) to express the answer in x.
Definite Integrals
Change the limits of integration when u-substituting in a definite integral: if u = g(x), the new limits are u = g(a) and u = g(b). Do not back-substitute — evaluate directly in u.
Worked Example
∫6x²(x³+1)⁴ dx
u = x³+1, du = 3x² dx
= 2∫u⁴ du = 2u⁵/5 + C
= 2(x³+1)⁵/5 + C
Integration by parts is the product rule in reverse. Use it when the integrand is a product of two different types of functions.
Choose u to be whichever function type appears earliest in the LIATE hierarchy. The remaining factor becomes dv.
L
Logarithmic
ln x, log₂x
I
Inverse Trig
arctan x, arcsin x
A
Algebraic
x², 3x+1, √x
T
Trigonometric
sin x, cos x, tan x
E
Exponential
eˣ, 5ˣ
∫x ln x dx (L before A → u = ln x)
dv = x dx → v = x²/2
= (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C
∫x eˣ dx (A before E → u = x)
dv = eˣ dx → v = eˣ
= xeˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C
∫arctan x dx (I before A → u = arctan x)
dv = dx → v = x
= x arctan x − ∫x/(1+x²) dx = x arctan x − (1/2) ln(1+x²) + C
When you need integration by parts multiple times and u is a polynomial, the tabular method is dramatically faster. Create a D column (differentiate u until 0) and an I column (integrate dv repeatedly). Multiply diagonally with alternating signs.
Example: ∫x³ eˣ dx
| Sign | D (differentiate) | I (integrate) |
|---|---|---|
| + | x³ | eˣ |
| − | 3x² | eˣ |
| + | 6x | eˣ |
| − | 6 | eˣ |
| + | 0 | eˣ |
Answer: x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C
These strategies handle integrals of the form ∫sinᵐx cosⁿx dx and ∫tanᵐx secⁿx dx by splitting off one factor and applying a Pythagorean identity.
m odd (sin power is odd)
Save one sin x, convert the rest via sin²x = 1 − cos²x, substitute u = cos x.
n odd (cos power is odd)
Save one cos x, convert the rest via cos²x = 1 − sin²x, substitute u = sin x.
Both m and n even
Use half-angle identities: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2. Reduces to integrals of cos 2x.
Key identity: tan²x = sec²x − 1. Use it to convert between powers of tan and sec.
n even (sec power is even)
Save sec²x, convert the rest via sec²x = 1 + tan²x, substitute u = tan x.
m odd (tan power is odd)
Save sec x tan x, convert remaining tan²x using tan²x = sec²x − 1, substitute u = sec x.
Special cases
∫tan x dx = ln|sec x| + C, ∫sec x dx = ln|sec x + tan x| + C (memorize these).
Trig substitution eliminates square roots of the form √(a²−x²), √(a²+x²), or √(x²−a²) by converting them to a trig expression without a radical. After integrating, use a right triangle to convert back to x.
√(a² − x²)
Sub: x = a sin θ
dx = a cos θ dθ
→ √(a² − x²) = a cos θ
Range: −π/2 ≤ θ ≤ π/2
Opposite = x, Hyp = a, Adj = √(a²−x²)
√(a² + x²)
Sub: x = a tan θ
dx = a sec²θ dθ
→ √(a² + x²) = a sec θ
Range: −π/2 < θ < π/2
Opposite = x, Adj = a, Hyp = √(a²+x²)
√(x² − a²)
Sub: x = a sec θ
dx = a sec θ tan θ dθ
→ √(x² − a²) = a tan θ
Range: 0 ≤ θ < π/2 or π ≤ θ < 3π/2
Hyp = x, Adj = a, Opposite = √(x²−a²)
∫ 1 / √(9 − x²) dx
a = 3, form √(a²−x²) → x = 3 sin θ, dx = 3 cos θ dθ
√(9 − 9 sin²θ) = 3 cos θ
∫ (3 cos θ dθ) / (3 cos θ) = ∫ dθ = θ + C
Back-substitute: θ = arcsin(x/3)
Answer: arcsin(x/3) + C
Partial fractions decomposes a rational function into simpler fractions whose antiderivatives are known (logarithms and arctangents). Use it whenever the integrand is a ratio of polynomials.
Check degree — divide if necessary
If the degree of the numerator ≥ degree of the denominator, perform polynomial long division first. Continue with the remainder only.
Factor the denominator completely
Factor Q(x) into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c with b² − 4ac < 0). Repeated factors require multiple terms.
Write the partial fraction form
Each distinct linear factor (ax+b) gets A/(ax+b). Each repeated factor (ax+b)ⁿ gets A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ. Each irreducible quadratic gets (Ax+B)/(ax²+bx+c).
Solve for the constants
Clear the denominators and either (a) substitute convenient values of x to isolate constants, or (b) expand and match coefficients of like powers of x.
Integrate each piece
∫A/(ax+b) dx = (A/a) ln|ax+b| + C. Irreducible quadratics with (Ax+B) in the numerator split into a logarithm part and an arctangent part via completing the square.
Example: ∫ (2x + 1) / (x² − x − 2) dx
Factor: x² − x − 2 = (x−2)(x+1)
(2x+1)/((x−2)(x+1)) = A/(x−2) + B/(x+1)
Multiply through: 2x+1 = A(x+1) + B(x−2)
x = 2: 5 = 3A → A = 5/3
x = −1: −1 = −3B → B = 1/3
∫ [5/(3(x−2)) + 1/(3(x+1))] dx
= (5/3) ln|x−2| + (1/3) ln|x+1| + C
Quick Reference: Decomposition Forms
| Factor type | Partial fraction form |
|---|---|
| Distinct linear: (ax+b) | A/(ax+b) |
| Repeated linear: (ax+b)² | A/(ax+b) + B/(ax+b)² |
| Repeated linear: (ax+b)ⁿ | A₁/(ax+b) + … + Aₙ/(ax+b)ⁿ |
| Irreducible quadratic: (ax²+bx+c) | (Ax+B)/(ax²+bx+c) |
| Repeated quadratic: (ax²+bx+c)² | (Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)² |
Improper integrals involve either infinite limits of integration (Type I) or integrands with vertical asymptotes within the interval (Type II). Both are evaluated using limits.
One or both limits of integration are ±∞. Replace the infinite limit with a variable and take a limit.
Definitions:
∫ₐ^∞ f(x) dx = lim_(b→∞) ∫ₐ^b f(x) dx
∫₋∞^b f(x) dx = lim_(a→−∞) ∫ₐ^b f(x) dx
∫₋∞^∞ f(x) dx = ∫₋∞^c f dx + ∫c^∞ f dx
The integrand blows up at some point c in [a, b]. Split the integral at c and take one-sided limits.
Definitions:
If discontinuity at b: lim_(t→b⁻) ∫ₐ^t f(x) dx
If discontinuity at a: lim_(t→a⁺) ∫t^b f(x) dx
If discontinuity at interior c: split at c, evaluate both sides
An improper integral converges if the limit exists and is a finite number. It diverges if the limit is ±∞ or does not exist.
p-Test (∫₁^∞ 1/xᵖ dx)
Converges if p > 1, diverges if p ≤ 1.
p = 2: ∫₁^∞ 1/x² dx = 1 (converges)
p = 1: ∫₁^∞ 1/x dx = ∞ (diverges)
p = 1/2: ∫₁^∞ 1/√x dx = ∞ (diverges)
p-Test (∫₀^1 1/xᵖ dx)
Converges if p < 1, diverges if p ≥ 1.
p = 1/2: ∫₀^1 1/√x dx = 2 (converges)
p = 1: ∫₀^1 1/x dx = ∞ (diverges)
p = 2: ∫₀^1 1/x² dx = ∞ (diverges)
Type I: ∫₁^∞ e^(−x) dx
= lim_(b→∞) [−e^(−x)]₁^b = lim_(b→∞) (−e^(−b) + e^(−1))
= 0 + 1/e = 1/e Converges.
Type II: ∫₀^1 1/√x dx
= lim_(t→0⁺) ∫t^1 x^(−1/2) dx = lim_(t→0⁺) [2√x]t^1
= lim_(t→0⁺) (2 − 2√t) = 2 − 0
= 2 Converges.
Type I (diverges): ∫₁^∞ 1/x dx
= lim_(b→∞) [ln x]₁^b = lim_(b→∞) ln b = ∞
Diverges.
Work through these questions in order. The first matching criterion determines your approach.
1. Does it match a basic antiderivative?
Yes → use the table directly. Done.
Check for xⁿ, eˣ, trig functions, 1/x, and inverse trig forms.
2. Can you simplify algebraically?
Yes → expand, factor, split the fraction, or apply a trig identity first.
Example: ∫sin 2x dx → use sin 2x = 2 sin x cos x or u = 2x.
3. Is there a composite function with its derivative nearby?
Yes → use u-substitution.
The derivative of the inner function must appear as a factor (up to a constant).
4. Is the integrand a product of two different function types?
Yes → use integration by parts. Apply LIATE to choose u.
If you need IBP more than once and u is a polynomial, switch to the tabular method.
5. Does it involve sinᵐ cosⁿ or tanᵐ secⁿ?
Yes → use trig integral strategies (odd power split or half-angle identities).
Check parity of each power to pick the right sub: u = cos x or u = sin x.
6. Does the integrand contain √(a²−x²), √(a²+x²), or √(x²−a²)?
Yes → use trigonometric substitution (sin, tan, or sec substitution).
Complete the square first if needed to get the integral into standard form.
7. Is it a rational function (polynomial / polynomial)?
Yes → use partial fraction decomposition.
Long-divide first if numerator degree ≥ denominator degree.
8. Is there an infinite limit or a vertical asymptote inside the interval?
Yes → it is an improper integral. Rewrite as a limit and evaluate.
If both endpoints are problematic, split the integral at a convenient interior point.
Let u = x² + 4, then du = 2x dx, so x dx = du/2
∫ x/(x²+4) dx = (1/2) ∫ 1/u du = (1/2) ln|u| + C
= (1/2) ln(x² + 4) + C
Note: x² + 4 > 0 always, so no absolute value needed.
x² is Algebraic, sin x is Trig → LIATE: u = x². Use tabular method.
D: x², 2x, 2, 0 I: sin x, −cos x, −sin x, cos x
Signs: +, −, +, −
= x²(−cos x) − 2x(−sin x) + 2(cos x) + C
= −x² cos x + 2x sin x + 2 cos x + C
m = 5 is odd → save one sin x, convert the rest using sin²x = 1 − cos²x
= ∫ (1 − cos²x)² sin x dx
Let u = cos x, du = −sin x dx
= −∫ (1 − u²)² du = −∫ (1 − 2u² + u⁴) du
= −u + (2/3)u³ − (1/5)u⁵ + C
= −cos x + (2/3) cos³x − (1/5) cos⁵x + C
a = 2, form √(a²−x²) → x = 2 sin θ, dx = 2 cos θ dθ
√(4 − 4 sin²θ) = 2 cos θ
∫ 2 cos θ · 2 cos θ dθ = 4 ∫ cos²θ dθ
= 4 ∫ (1 + cos 2θ)/2 dθ = 2θ + sin 2θ + C
= 2θ + 2 sin θ cos θ + C
Back-sub: θ = arcsin(x/2), sin θ = x/2, cos θ = √(4−x²)/2
= 2 arcsin(x/2) + (x/2)√(4−x²) + C
Factor: x² − 1 = (x−1)(x+1)
1/((x−1)(x+1)) = A/(x−1) + B/(x+1)
1 = A(x+1) + B(x−1)
x = 1: 1 = 2A → A = 1/2 x = −1: 1 = −2B → B = −1/2
∫ [(1/2)/(x−1) − (1/2)/(x+1)] dx
= (1/2) ln|x−1| − (1/2) ln|x+1| + C = (1/2) ln|(x−1)/(x+1)| + C
Use u-substitution when you can identify a composite function inside the integral and its derivative (or a constant multiple of it) also appears in the integrand. Look for a function and its derivative together: ∫f(g(x))g'(x) dx. Set u = g(x) (the inner function), compute du = g'(x) dx, and substitute to simplify. Example: ∫2x·cos(x²) dx — let u = x², du = 2x dx, transforms to ∫cos(u) du = sin(u) + C = sin(x²) + C.
LIATE is a mnemonic for choosing u in integration by parts (∫u dv = uv − ∫v du). Choose u to be the type of function that appears earliest in the acronym: L = Logarithmic (ln x, log x), I = Inverse trig (arctan x, arcsin x), A = Algebraic (polynomials like x², 3x + 1), T = Trigonometric (sin x, cos x), E = Exponential (eˣ, 2ˣ). Example: ∫x·eˣ dx — x is Algebraic and eˣ is Exponential; A comes before E, so u = x and dv = eˣ dx. Result: xeˣ − eˣ + C.
Use the tabular method (also called the DI method or repeated integration by parts) when you must apply integration by parts multiple times and u is a polynomial. Create two columns: D (differentiate u repeatedly until reaching 0) and I (integrate dv repeatedly). Alternate the signs +, −, +, −, … and multiply diagonally. Example: ∫x³eˣ dx — D column: x³, 3x², 6x, 6, 0; I column: eˣ, eˣ, eˣ, eˣ, eˣ. Answer: x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C.
For ∫sinᵐx cosⁿx dx: If m is odd, save one sin x, convert the rest using sin²x = 1 − cos²x, then substitute u = cos x. If n is odd, save one cos x, convert the rest using cos²x = 1 − sin²x, then substitute u = sin x. If both m and n are even, use half-angle identities: sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. Example: ∫sin³x dx = ∫sin²x · sin x dx = ∫(1 − cos²x) sin x dx; let u = cos x, du = −sin x dx → −∫(1 − u²) du = −cos x + cos³x/3 + C.
The three trig substitutions match the three Pythagorean identities. (1) For √(a² − x²): substitute x = a sin θ, dx = a cos θ dθ. The expression becomes a cos θ. Use when the integrand contains a² − x². (2) For √(a² + x²): substitute x = a tan θ, dx = a sec² θ dθ. The expression becomes a sec θ. Use when the integrand contains a² + x². (3) For √(x² − a²): substitute x = a sec θ, dx = a sec θ tan θ dθ. The expression becomes a tan θ. Use when the integrand contains x² − a². Always draw a right triangle to convert back to x at the end.
Partial fraction decomposition rewrites a rational function P(x)/Q(x) as a sum of simpler fractions that are easy to integrate. Step 1: If degree of P ≥ degree of Q, perform polynomial long division first. Step 2: Factor the denominator Q(x) completely. Step 3: Write a partial fraction for each factor — A/(ax+b) for linear factors, (Ax+B)/(ax²+bx+c) for irreducible quadratics, and repeated factors require extra terms like A/(ax+b) + B/(ax+b)². Step 4: Clear denominators and solve for the constants by substituting convenient values of x or matching coefficients. Step 5: Integrate each partial fraction — linear factors give logarithms, irreducible quadratics give arctangents.
An improper integral has either an infinite limit of integration (Type I) or an integrand with a vertical asymptote in the interval (Type II). To evaluate, replace the problematic part with a limit. Type I example: ∫₁^∞ 1/x² dx = lim_{b→∞} ∫₁^b x⁻² dx = lim_{b→∞} [−1/x]₁^b = lim_{b→∞} (−1/b + 1) = 1. This converges. Type II example: ∫₀^1 1/√x dx = lim_{a→0⁺} ∫ₐ^1 x^{−1/2} dx = lim_{a→0⁺} [2√x]ₐ^1 = 2. This converges. If the limit does not exist or is ±∞, the integral diverges. The p-test: ∫₁^∞ 1/xᵖ dx converges if p > 1 and diverges if p ≤ 1.
Follow this decision process: (1) Is it a basic form? Check the antiderivative table first — many integrals are done immediately. (2) Can you simplify? Expand products, split fractions, use trig identities before trying a technique. (3) Is there a composite function with its derivative? Use u-substitution. (4) Is it a product of two different function types? Use integration by parts — apply LIATE to choose u. (5) Does it involve sin^m · cos^n, tan^m · sec^n, etc.? Use trig integral strategies. (6) Does it contain √(a²−x²), √(a²+x²), or √(x²−a²)? Use trig substitution. (7) Is it a rational function (polynomial over polynomial)? Use partial fractions. (8) Does it have an infinite bound or vertical asymptote? It is an improper integral — convert to a limit.
Reduction formulas reduce the power of a trig function by 2 each time, eventually arriving at a base case. For sine: ∫sinⁿx dx = −(1/n)sinⁿ⁻¹x cos x + ((n−1)/n)∫sinⁿ⁻²x dx. For cosine: ∫cosⁿx dx = (1/n)cosⁿ⁻¹x sin x + ((n−1)/n)∫cosⁿ⁻²x dx. For secant (very useful in Calc 2): ∫secⁿx dx = (1/(n−1))secⁿ⁻²x tan x + ((n−2)/(n−1))∫secⁿ⁻²x dx. Apply the formula repeatedly until reaching ∫sin x dx, ∫cos x dx, ∫sec x dx, or ∫sec²x dx — all of which are standard antiderivatives.
Chain rule, product rule, quotient rule, and implicit differentiation — the foundation for integration
L'Hôpital's rule, continuity, and the formal definition of the derivative and integral
Convergence tests, power series, Taylor and Maclaurin series, radius of convergence
Interactive problems with step-by-step solutions and private tutoring — identify your weakest technique and drill it to mastery.
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