Indefinite Integrals

01 / UNDERSTAND & APPLY

1. An indefinite integral is a family of functions

An antiderivative of f on an interval I is a differentiable function F with F′(x)=f(x) for all interior points of I. The notation ∫f(x)dx denotes the antiderivative family F(x)+C, not a single numerical area. The variable after d identifies the variable of integration.

Any two antiderivatives on an interval differ by a constant. Their difference has derivative zero, so the mean value theorem makes it constant on that interval. The interval condition matters: on a disconnected domain, different connected components may carry independent constants.

For example, ∫dx/x=ln|x|+C on any interval not containing zero. If describing all antiderivatives on ℝ∖{0}, the constant on x<0 need not equal the constant on x>0. Writing one C is conventional shorthand for an interval-wise answer, not a constraint joining disconnected pieces.

Integration reverses differentiation but does not provide a universal elementary formula for every integrand. A continuous function has an antiderivative constructed by a definite integral, yet that antiderivative may not be expressible through a finite combination of elementary functions. Failure of a familiar technique does not prove that no antiderivative exists.

02 / UNDERSTAND & APPLY

2. Recognize derivative patterns before choosing machinery

Linearity gives ∫(af+bg)dx=a∫f dx+b∫g dx for constants a,b, with the final arbitrary constants absorbed into one constant on the interval. It does not give a rule for integrating a product as the product of integrals or a quotient as their quotient.

The power pattern is ∫xʳdx=xʳ⁺¹/(r+1)+C for r≠−1 on a domain where the real power and derivative are valid. For arbitrary real r, x>0 is a safe setting; integer powers allow their usual larger domains. The exceptional exponent −1 produces a logarithm rather than division by zero.

Useful basic patterns include ∫eˣdx=eˣ+C, ∫cos x dx=sin x+C, ∫sin x dx=−cos x+C, ∫sec²x dx=tan x+C and ∫dx/(1+x²)=arctan x+C. Trigonometric arguments are in radians and the integrand’s poles remain excluded.

Linear inner functions introduce reciprocal scale factors: for a≠0, ∫e^(ax+b)dx=e^(ax+b)/a+C. A quick derivative of the proposed result checks that the chain-rule factor a is canceled. If a=0, integrate the resulting constant directly instead of using an invalid formula.

03 / UNDERSTAND & APPLY

3. Substitution is the chain rule read backwards

If F′(u)=f(u), then d[F(g(x))]/dx=f(g(x))g′(x). Therefore an integral of f(g(x))g′(x) becomes ∫f(u)du after setting u=g(x). The differential notation du=g′(x)dx records the factor that must be present; it does not permit treating dx as an unrelated disposable symbol.

A successful substitution converts the entire integrand and differential to the new variable. Leaving a mixture of x and u without an explicit relation is incomplete. Constant mismatches are easy to correct, but a missing nonconstant factor cannot be inserted for free.

The inner derivative may be hidden after algebraic rearrangement. If u=x²+1, then x dx=du/2 and x²=u−1. This lets an integrand containing x³dx be rewritten as (u−1)du/2. Identifying the inner expression is only the first step; every remaining factor must also be transformed.

For indefinite integrals, return to the original variable at the end. A substitution need not be globally one-to-one when it is justified directly by the chain rule, but solving x in terms of u or introducing inverse-trigonometric branches may require restricting to a suitable interval.

04 / UNDERSTAND & APPLY

4. A denominator and its derivative suggest a split

For a rational expression with a quadratic denominator, compare the numerator with the derivative of that denominator. A matching component integrates to a logarithm. Any remaining constant term can often be handled by completing the square and recognizing an inverse tangent or a logarithmic partial fraction.

For Q(x)=ax²+bx+c with a≠0, write a linear numerator mx+n as A Q′(x)+B by matching coefficients. This is an algebraic identity, so it does not approximate the integrand. Then integrate A Q′/Q and B/Q separately on intervals where Q≠0.

The sign of the completed-square constant matters. A positive square sum leads to an arctangent pattern; a difference of squares leads to real linear factors and logarithms; a repeated square leads to a reciprocal-power pattern. Do not use an arctangent formula solely because a quadratic appears.

05 / UNDERSTAND & APPLY

5. Parts transfers a derivative to a better factor

The product rule gives (uv)′=u′v+uv′. Rearranging and integrating yields ∫u dv=uv−∫v du. The method is useful when differentiating one factor simplifies it and integrating the other is feasible. The new integral should be easier or should connect back to the original in a solvable way.

For a polynomial times an exponential or sine/cosine, repeated differentiation of the polynomial eventually removes it. For a logarithm or inverse-trigonometric function by itself, regard the integrand as that function times 1 and choose dv=dx. Familiar priority mnemonics can suggest a choice, but the algebraic outcome is the real test.

Choose one convenient antiderivative v for dv during the calculation; additional constants inside intermediate v expressions ultimately cancel or merge into the final C. Carrying a fresh unrelated constant through every line obscures the structure.

For cyclic integrals such as eˣcos x, two applications of parts return the original integral with a coefficient. Move that term to the other side and solve the resulting algebraic equation. The recurrence is useful, not evidence that the method has failed.

06 / UNDERSTAND & APPLY

6. Rational functions need the right decomposition

For P(x)/Q(x), first divide polynomials if the numerator degree is at least the denominator degree. Partial fractions is then applied to the proper rational remainder. The polynomial quotient is integrated separately.

Factor the real denominator into linear factors and irreducible quadratics. A simple linear factor x−a contributes A/(x−a). A repeated factor (x−a)ᵏ needs every power A₁/(x−a)+…+Aₖ/(x−a)ᵏ. Omitting lower powers can make the decomposition impossible even when the method is appropriate.

For an irreducible quadratic factor Q, use a numerator Ax+B over each required power of Q. A constant numerator alone is not general enough. Solve the coefficient identity by convenient substitutions and coefficient comparison, then verify the decomposition before integrating.

The decomposition is valid away from denominator zeros. Logarithms of linear factors require absolute values for an interval-wise real formula. Powers above the first integrate by the power rule, so repeated factors do not produce only logarithms.

07 / UNDERSTAND & APPLY

7. Identities should create a derivative pair

For products of powers of sine and cosine, an odd sine power suggests reserving one sin x dx and converting the remaining even power with sin²x=1−cos²x. Then u=cos x completes a substitution. An odd cosine power similarly suggests u=sin x.

When both powers are even, use half-angle identities sin²x=(1−cos2x)/2 and cos²x=(1+cos2x)/2 to reduce them. Products at different frequencies can be converted to sums. Every frequency coefficient introduced by an identity must be handled when integrating.

For powers of tangent and secant, an even secant power can reserve sec²x dx for u=tan x, using sec²x=1+tan²x. An odd tangent power with a secant factor can suggest u=sec x. These are useful patterns, not exhaustive rules; some cases, such as sec³x, use integration by parts.

Trigonometric antiderivatives hold on intervals where the integrand exists. Two equivalent-looking answers may differ by constants on separate intervals, especially after inverse trigonometric or tangent-half-angle substitutions. Differentiate on the specified interval to check equivalence.

08 / UNDERSTAND & APPLY

8. Choose a branch before simplifying a square root

Radicals involving a²−x², a²+x² and x²−a² often suggest x=a sin θ, x=a tan θ and x=a sec θ, respectively, with a>0. The corresponding identities simplify the radical, but square roots produce absolute values unless the chosen θ interval fixes the sign.

For √(a²−x²), choose −π/2<θ<π/2 with x=a sin θ when |x|<a. Then cos θ>0 and the radical equals a cos θ, while dx=a cos θ dθ. Both the radical and the differential must be transformed.

For √(a²+x²), the tangent substitution on the same θ interval makes sec θ positive. For √(x²−a²), the real domain has separate positive and negative branches; a secant substitution must be chosen and interpreted separately on an appropriate interval. Ignoring |sec θ| can flip a sign.

After integration, return to x using the chosen branch and algebraic identities. A right-triangle sketch can help for positive quantities, but it does not justify dropping signs on negative intervals. The derivative check is the final safeguard.

09 / UNDERSTAND & APPLY

9. A half-angle substitution can convert the whole integrand

For a rational expression in sin x and cos x, the substitution t=tan(x/2) gives sin x=2t/(1+t²), cos x=(1−t²)/(1+t²) and dx=2dt/(1+t²). Thus the integral becomes rational in t wherever the substitution is defined.

The substitution excludes x=(2k+1)π, even when the original integrand is defined there. The resulting inverse-tangent antiderivative is therefore initially an interval-wise formula. To extend across an artificial substitution break, choose constants so that neighboring branches join appropriately.

This general method may be longer than a direct identity or a simple derivative substitution. Use it when the whole rational trigonometric structure benefits from conversion, rather than automatically applying it to every sine or cosine integral.

10 / UNDERSTAND & APPLY

10. Choose the simplest reversible structure

First simplify algebraically while preserving the original domain. Then look for an inner expression and its derivative. If the integrand is a product with a simplifying derivative, consider parts. If it is rational, divide and factor before decomposing. If it is trigonometric or radical, look for an identity that creates a recognizable substitution.

A recurrence or a second appearance of the original integral can be productive. For example, repeated parts may yield I=H−I, so I=H/2. Track coefficients carefully and include only the final arbitrary constant rather than treating every intermediate antiderivative equality as a fixed numerical equation.

Differentiate the final expression completely and simplify to the original integrand on the stated domain. This checks missing chain factors, signs, constants multiplying inverse functions and partial-fraction coefficients. Two answers with the same derivative on an interval differ only by a constant, even if their appearances differ.

An initial value F(x₀)=y₀ can determine C on the connected interval containing x₀. It does not determine constants on other components separated by poles. Do not carry an antiderivative through an integrand singularity as if the function were defined there.

11 / UNDERSTAND & APPLY

11. Original practice with full solutions

Give an antiderivative on an interval where the integrand is defined. All constants below are arbitrary unless an initial value is supplied.

12 / UNDERSTAND & APPLY

12. A verified antiderivative is the finish line

The most reliable integration skill is recognizing the derivative structure concealed by notation. Substitution reverses the chain rule; parts reverses the product rule; partial fractions separates rational structures into known pieces; trigonometric identities reshape powers into derivative pairs.

Common mistakes include forgetting C, using the power rule at exponent −1, inserting a missing variable factor in substitution, omitting repeated partial-fraction terms, losing an inner scale factor and dropping absolute values from logarithms or radicals without checking signs.

Keep domains attached to the work. A pole separates intervals, a square root fixes a real branch, and a tangent-half-angle substitution can introduce a break not present in the original integrand. These are mathematical conditions, not optional footnotes.

Finally differentiate the answer. If the derivative agrees with the integrand throughout the intended interval, the antiderivative is established. If a preferred elementary technique fails, reconsider the structure without assuming that every continuous function must have an elementary antiderivative.