Integration by parts and other techniques

Two changing factors make two derivative terms

A rectangle has sides u=2 and v=3. Its sides grow to 3 and 5, so its area grows from 6 to 15. The added area has three pieces: a strip of area 4, a strip of area 3 and a corner of area 2. Multiplying the side changes alone gives only that corner, not the whole area change.

This distinction matters when a product changes continuously. Take u(x)=x and v(x)=x². Their product is x³, whose rate of change is 3x². A prime marks a derivative, the rate of change with respect to x. One part comes from the change of v, giving u v′=2x²; the other comes from the change of u, giving v u′=x². These two terms add to 3x².

On the interval from 0 to 2, the product increases by 8. The two signed accumulations are 16/3 and 8/3, adding to that change. We can find the first by taking the product change and subtracting the second. This is integration by parts. For this polynomial there is also a shorter power-rule calculation; the example makes the exchange visible before using it on a harder product.

For preparation, revisit Functions for original domains and values, Expanding and factoring for equivalent expressions, and Fraction operations for signed sums and common denominators. The needed calculus notation is defined below. First follow the polynomial and rational examples; return to the product-rule proof and the optional exponential, logarithmic and trigonometric examples when their derivative rules are familiar.

Check a proposed integral by differentiating it

For a real function f on an open interval I, an antiderivative is a differentiable function F on I satisfying F′(x)=f(x) at every input of I. The derivative F′(x) is the finite limit of [F(x+h)−F(x)]/h as nonzero h tends to zero inside the interval. An indefinite integral ∫f(x)dx denotes the family of antiderivatives, not a numerical sum over an unspecified interval.

If one antiderivative is F, every F+C is another because a constant has derivative zero. Conversely, two antiderivatives on the same interval differ by a constant: their difference has derivative zero, and the mean value theorem gives zero change between any two inputs. The theorem says that a function continuous on a closed interval and differentiable inside has endpoint change equal to an interior derivative times the input length. Separate disconnected intervals can have different constants.

Here are the derivative facts used in the calculations: (xⁿ)′=nxⁿ⁻¹ for positive integers n, (eˣ)′=eˣ, (ln x)′=1/x for x>0, (sin x)′=cos x and (cos x)′=−sin x for angles in radians. The chain rule gives (G(g(x)))′=G′(g(x))g′(x) when both derivatives exist. These standard elementary-function rules are the starting facts for the optional nonpolynomial examples; the product exchange is derived separately below.

Thus ∫xⁿdx=xⁿ⁺¹/(n+1)+C for nonnegative integers n; ∫eᵏˣdx=eᵏˣ/k+C when k≠0; and ∫cos(kx)dx=sin(kx)/k+C when k≠0. At k=0 the integrands are constant 1, so their primitive is x+C. The reciprocal rule is ∫1/(x−c)dx=ln|x−c|+C on an interval excluding c. Differentiate on each side to verify the absolute value: both sides give 1/(x−c).

Move one term of the product rule to the other side

Write Δu=u(x+h)−u(x) and Δv=v(x+h)−v(x). The exact finite identity is Δ(uv)=uΔv+vΔu+ΔuΔv, with u and v evaluated at x. Divide by h. If u and v have finite derivatives, then Δu/h tends to u′, Δv/h tends to v′, and Δv tends to zero. The corner contribution (Δu/h)Δv therefore tends to zero. This proves (uv)′=u v′+v u′; the corner is not discarded for a finite change.

Assume u and v are continuously differentiable on an open interval I, meaning that their derivatives exist and are continuous there. Let H be any antiderivative of v u′ on I. Then (uv−H)′=u v′+v u′−v u′=u v′. Hence ∫u v′dx=uv−∫v u′dx, with one final arbitrary constant. The right-hand integral means a chosen primitive before that final constant is added.

The compact notation ∫u dv=uv−∫v du abbreviates the same identity: du=u′dx and dv=v′dx. These differential labels record derivative factors; they are not instructions to multiply two separate indefinite integrals. Choose v with v′ equal to the whole factor assigned to dv, including any constant multiplier.

Adding a constant K to the chosen v changes uv by Ku and changes a chosen primitive of v u′ by the matching Ku. They cancel up to the final arbitrary constant. Taking a convenient primitive v is therefore legitimate; carrying several independent constants through intermediate lines usually obscures this cancellation.

The corner has area ΔuΔv. After division by the input increment its contribution tends to zero when both derivatives exist. It is not zero for this finite change.

Integrate one factor and simplify the other

For ∫x eˣdx, choose u=x and dv=eˣdx. Then du=dx and v=eˣ. The exchange gives x eˣ−∫eˣdx=(x−1)eˣ+C. Differentiate the proposed answer: eˣ+(x−1)eˣ=x eˣ. The new integral is easier because differentiating x removes it.

Choosing u=eˣ and dv=x dx is algebraically allowed, but it leaves an integral containing x²eˣ/2. It has made the polynomial factor larger instead of smaller. A good choice needs both an accessible primitive for dv and a useful remaining integral. A mnemonic can suggest candidates but does not prove that a choice helps.

For ∫ln x dx with x>0, regard the second factor as 1: u=ln x, dv=dx, du=dx/x and v=x. The result is x ln x−x+C. Here a factor is created deliberately, and the condition x>0 remains in the answer. The real logarithm is not defined at negative inputs.

A substitution fits a derivative of an inner expression: ∫2x cos(x²)dx=sin(x²)+C follows directly by the chain rule. The product ∫x cos x dx instead fits parts with u=x and v=sin x, giving x sin x+cos x+C. Do not choose a method from the appearance of a multiplication sign alone; inspect the actual derivative pattern.

A second exchange can reduce a power or return the original

For ∫x²eˣdx, parts first gives x²eˣ−2∫x eˣdx. Substitute the already checked primitive to obtain eˣ(x²−2x+2)+C. Its derivative is eˣ[(x²−2x+2)+(2x−2)]=x²eˣ. Polynomial degree decreases one step at a time when the polynomial is differentiated.

For I=∫eˣcos x dx, choose u=cos x and dv=eˣdx. Then I=eˣcos x+J, where J denotes a chosen primitive of eˣsin x. Applying parts to J gives J=eˣsin x−I up to constants. Combine to obtain 2I=eˣ(cos x+sin x) up to a constant, so I=eˣ(cos x+sin x)/2+C.

A returned integral is not always a failure, but it must produce an equation with a nonzero coefficient of the unknown primitive. A mere identity I=I supplies no new calculation. Check the solved answer by differentiating; here the sine terms cancel and the two cosine terms combine to eˣcos x.

Why a derivative gives an endpoint difference

A definite Riemann integral ∫ₐᵇf(x)dx for a<b is the common finite limit of sums Σf(ξᵢ)Δxᵢ over partitions of [a,b] as the largest positive cell width tends to zero, for every choice of tag ξᵢ in its cell. Continuity on a finite closed interval is sufficient for this limit: uniform continuity makes the maximum height variation in every sufficiently narrow cell small, and the lower–upper sum gap is bounded by that variation times b−a.

For continuous f on an open interval I and a fixed c∈I, put A(x)=∫꜀ˣf(t)dt with the usual signed orientation. Additivity gives [A(x+h)−A(x)]/h=(1/h)∫ₓˣ⁺ʰf(t)dt. Subtract f(x); its magnitude is bounded by the largest |f(t)−f(x)| between x and x+h. Continuity makes this tend to zero, also for negative h. Thus A′=f.

If F′=f on I, the zero-derivative argument makes F−A constant. Consequently ∫ₐᵇf=F(b)−F(a) whenever a and b lie in I. This supplies the local fundamental-theorem bridge used here. Products u v′ and v u′ are continuous under our continuously differentiable hypotheses, so their integrals exist on every closed subinterval of I.

Integrating the product rule therefore gives ∫ₐᵇu v′dx=u(b)v(b)−u(a)v(a)−∫ₐᵇv u′dx. The notation [uv]ₐᵇ means upper value minus lower value, including both endpoints. Define an integral with equal endpoints to be 0, and reversing endpoints to negate the forward integral. The same parts identity then holds for either orientation.

The boundary term is a change, not just a final value

For ∫₀¹x eˣdx, the checked primitive gives [(x−1)eˣ]₀¹=0−(−1)=1. The parts form gives [x eˣ]₀¹−∫₀¹eˣdx=e−(e−1)=1. Ignoring the lower endpoint would lose the answer. A positive integrand with reversed endpoints instead gives −1.

If the input is time measured in seconds, u has units metres and v has units newtons, then v′ has units newtons per second. Each of ∫u v′dt, ∫v u′dt and [uv] has units newton metres. In ∫v u′dt, u′ is metres per second, so v u′ can be power and its time integral work. The product identity needs signed values and rates; identifying a quantity as physical work also needs the relevant force-and-displacement model.

A plotted height below zero contributes negatively on a forward interval. Geometric area would integrate the magnitude instead. Integration by parts preserves the signed integral; it does not turn cancellation into a sum of absolute areas. Preserve the same bounds in the remaining integral and evaluate every term with compatible units.

A sum of simple fractions can reveal a primitive

A rational function is a quotient P(x)/Q(x) of polynomials on inputs where Q(x)≠0. For (2x+1)/((x−1)(x+2)), try A/(x−1)+B/(x+2). Taking a common denominator gives numerator A(x+2)+B(x−1). To equal 2x+1 for every input, solve A+B=2 and 2A−B=1, giving A=B=1.

Thus the quotient equals 1/(x−1)+1/(x+2) wherever x≠1 and x≠−2. The numerator equation is a polynomial identity, so substituting 1 or −2 into that equation to find coefficients is valid. It does not evaluate the original quotient at its excluded inputs.

An antiderivative is ln|x−1|+ln|x+2|+C on any one of the intervals (−∞,−2), (−2,1) or (1,∞). Its derivative is exactly the original function on that interval. Each interval can have its own constant; the displayed formula is not a primitive across either missing input.

For two distinct constants r and s, a linear numerator αx+β admits A/(x−r)+B/(x−s), because αx+β=A(x−s)+B(x−r) has a unique solution when r≠s. Setting x=r and x=s gives A=(αr+β)/(r−s) and B=(αs+β)/(s−r). The recombined polynomial identity proves the decomposition. This two-factor proof does not use the same formula when the factors coincide.

Separate a polynomial part before decomposing

When the numerator degree is at least the denominator degree, first use polynomial division: P=SQ+R, with R=0 or degree R smaller than degree Q. On Q≠0, the quotient is S+R/Q. The polynomial part and proper fraction then have separate integrals, with the original denominator exclusions retained.

For (x²+2x+3)/(x+1), the exact numerator identity is x²+2x+3=(x+1)(x+1)+2. Hence the quotient is x+1+2/(x+1), and a primitive is x²/2+x+2ln|x+1|+C on either interval excluding −1. Recombining the quotient and differentiating the primitive provide two different checks.

Polynomial division is also useful when a cancellation is possible. For (x²−1)/(x−1), the simplified expression is x+1 only on x≠1. An antiderivative on each original interval is x²/2+x+C. Choosing to extend the simplified function to 1 describes a different domain, so a definite integral crossing 1 is not covered by the continuous-domain rule stated here.

Different factors require different primitive patterns

A repeated factor needs more than one possible power: 1/[x(x−1)²]=1/x−1/(x−1)+1/(x−1)² on x≠0,1. Recombine the right-hand side to check the numerator 1. The resulting primitive is ln|x|−ln|x−1|−1/(x−1)+C on each original interval. The last term is not another logarithm, because the reciprocal-square integrand has a different power.

More generally, a repeated linear factor (x−c)ᵏ calls for terms A₁/(x−c),…,Aₖ/(x−c)ᵏ when setting up partial fractions. For j≥2, a primitive of (x−c)⁻ʲ is (x−c)¹⁻ʲ/(1−j); differentiating verifies it. For j=1 use ln|x−c| instead. Preserve x≠c in every case.

A real irreducible quadratic has no real zero and should not be split into invented real linear factors. Its partial-fraction numerator can need both Ax+B. For example, ∫(2x+3)/(x²+1)dx=ln(x²+1)+3 arctan x+C on all real inputs: the derivative rules are (ln(x²+1))′=2x/(x²+1) and (arctan x)′=1/(1+x²). Here arctan x is the unique angle in (−π/2,π/2) with tangent x, and its derivative follows by differentiating that inverse relationship with nonzero secant square. This is an optional additional elementary-function rule.

These examples explain the needed patterns; they do not supply a symbolic algorithm for every denominator. First prove the particular algebraic identity you use. If a proposed factorisation has complex roots or a denominator vanishes on the integration interval, keep that limitation visible rather than apply a real continuous-domain formula without its conditions.

A useful identity changes the form, not the values

For angles in radians, cos(2x)=1−2sin²x and sin²x+cos²x=1 give sin²x=(1−cos2x)/2 and cos²x=(1+cos2x)/2. Apply linearity and the chain factor: primitives are x/2−sin(2x)/4 and x/2+sin(2x)/4. Differentiating the sine term gives twice its cosine, so the coefficient must be 1/4 rather than 1/2.

Over [0,π], both squared functions have integral π/2 because sin(2π)=sin0=0. They are not the same function: their peaks occur at different inputs. A common definite integral does not prove equality of integrands. Reversing the endpoints gives −π/2, although both original functions are nonnegative.

For sin³x cos x, reserve the cos x factor as the derivative of sin x. Differentiating sin⁴x/4 yields sin³x cos x, so it is a primitive on every real interval. From 0 to π/2 the endpoint difference is 1/4. By contrast, sin²x has no matching cos x factor; the half-angle identity is the useful rewrite here.

Each rewrite must hold on the original domain. Sine and cosine are defined on all real inputs; identities involving tangent or a division by cosine would add exclusions if they were used. A trigonometric substitution for a radical is a different method and needs its own domain and inverse-branch checks; it is outside this lesson's worked scope.

Make the next step simpler, then verify the whole answer

Start by writing the domain and simplifying exact algebra where permitted. A visible derivative of an inner expression suggests a chain-pattern substitution. A product with a simpler derivative for one factor and an accessible primitive for the other suggests parts. A rational quotient suggests division when needed, followed by a proved partial-fraction identity. Trigonometric squares can suggest a power-reducing identity.

These are decision aids, not a claim that every continuous function has an elementary-looking antiderivative. The function e^(x²) is continuous and therefore has a local integral-defined primitive, but the fact that it is continuous does not provide a formula built from the elementary operations used here. Getting stuck with one factor choice does not prove that no primitive exists.

For an indefinite answer, differentiate it on every claimed original interval and compare with the whole integrand, including signs and constant multipliers. State one constant per connected interval. For a definite answer, also check existence on the entire closed interval, retain endpoint order and subtract the lower value. Numerical samples and a plausible graph can reveal a mistake; they do not replace an identity or an all-input argument.

Use the explorer to compare the exact contributions and deliberately test excluded inputs. Return to the local bridge if an endpoint subtraction lacks its continuity conditions, to the factor choice if the remaining integral became harder, and to the domain discussion if a simplification appears to fill a missing input.

Worked examples

Try it yourself

1. Is (∫x dx)(∫x dx)=x⁴/4 a primitive of x² when the intermediate constants are zero?

   Show the answer and reasoning

   No.

   The derivative of x⁴/4 is x³, not x². The product rule supplies two terms; multiplying separate primitives is not an integration rule.

2. For u=x² and v=x³ on [0,1], find ∫u v′, ∫v u′ and the product change.

   Show the answer and reasoning

   3/5, 2/5 and 1 respectively.

   The two integrands are 3x⁴ and 2x⁴, with primitives 3x⁵/5 and 2x⁵/5. Their endpoint differences add to [x⁵]₀¹=1.

3. If v is constant, must ∫u v′ on a finite interval equal the whole product change?

   Show the answer and reasoning

   No; the original integral is 0.

   Then v′=0, while the other term ∫v u′ equals [uv]. The subtraction in parts cancels this product change.

4. Find ∫x e²ˣdx.

   Show the answer and reasoning

   e²ˣ(x/2−1/4)+C.

   Choose u=x and v=e²ˣ/2. Subtract ∫e²ˣ/2 dx=e²ˣ/4, then differentiate the result to check the factor 2.

5. Does x ln x−x+C supply a real primitive of ln x at x=−1?

   Show the answer and reasoning

   No; this original function requires x>0.

   The separate function ln|x| is defined for x≠0 and has a related primitive on each side. Replacing ln x by ln|x| changes the original function's domain.

6. Find ∫₁²x dx, and explain why using only x²/2 at 2 is wrong.

   Show the answer and reasoning

   3/2.

   The endpoint difference is 2−1/2. The value at the lower endpoint cannot be omitted merely because a convenient primitive was chosen.

7. Find ∫_π^0 sin²x dx.

   Show the answer and reasoning

   −π/2.

   The forward integral is π/2. Reversal changes its sign rather than changing the nonnegative integrand into geometric area.

8. Can A/(x−1)+B/(x−1) represent 1/(x−1)² for every x≠1 with constant A,B?

   Show the answer and reasoning

   No; a reciprocal-square term is needed.

   The left side is (A+B)/(x−1). Multiplying by (x−1)² would require (A+B)(x−1)=1 for every allowed input, which no constant A+B satisfies.

9. May the local endpoint rule for continuous integrands be applied directly to ∫₀²1/(x−1) dx?

   Show the answer and reasoning

   No.

   The original function is undefined at an interior input and is not bounded on this closed interval. A formal logarithmic endpoint difference does not establish this Riemann integral. Improper limits are a separate topic.

10. On the domain x≠0, must a primitive of 1/x use the same constant on positive and negative inputs?

    Show the answer and reasoning

    No.

    Both ln|x|+C₊ on (0,∞) and ln|x|+C₋ on (−∞,0) differentiate to 1/x. The zero-derivative constant proof compares inputs within one connected interval.

11. Find ∫1/(x−2)³dx on an interval excluding 2.

    Show the answer and reasoning

    −1/[2(x−2)²]+C.

    The power rule with exponent −3 gives coefficient 1/(−2). Differentiating gives a positive reciprocal cube; x=2 remains excluded.

12. Which directly checked primitive fits 2x cos(x²): sin(x²) or x sin(x²)?

    Show the answer and reasoning

    sin(x²)+C.

    Its chain-rule derivative is cos(x²)·2x. The second candidate also creates a sin(x²) term and an extra factor x, so it is not a primitive of the stated integrand.

13. Both sin²x and cos²x integrate to π/2 on [0,π]. Does that prove they are equal functions?

    Show the answer and reasoning

    No.

    At x=0 their values are 0 and 1. Equality of one signed accumulation is much weaker than equality at every input.

14. If u is metres, v is newtons and the input is seconds, what units does ∫v u′dt have?

    Show the answer and reasoning

    Newton metres.

    u′ has units metres per second. Multiplying by newtons and integrating over seconds leaves newton metres, matching the product endpoint change.

What to carry forward

• Integration by parts rearranges the two terms of the product rule; it does not multiply separate primitives.
• Choose an accessible primitive for one factor and a simpler derivative for the other, then check the remaining integral.
• A local antiderivative family uses one arbitrary constant on each connected original interval.
• A definite calculation requires its existence conditions, both endpoint values, signed orientation and compatible units.
• Rational rewrites preserve original exclusions; division and factor patterns guide valid decompositions.
• A trigonometric identity or chain pattern must preserve every original value and its derivative multipliers.
• Differentiation and recombination establish the answer; finite examples and graphs illustrate it.