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Repeated integration by parts

Apply integration by parts repeatedly to polynomial products and logarithmic powers. Track coefficients and signs, solve cyclic integrals and check exact definite results.

Before you startIntegration by parts, product-rule differentiation and standard integrals.

01 / Keep going while the remainder simplifies

One application may reduce the problem without finishing it.

Iₙ = ∫xⁿeˣ dx = xⁿeˣ − nIₙ₋₁

Use the recurrence for a positive integer n; finish with the base integral I₀ = eˣ + C.

Each application differentiates the polynomial and integrates the exponential. Carry the full coefficient and sign outside the next integral. Stop only when the remainder is a standard integral; include one final constant of integration.

Reduce the polynomial one degree at a timeExplore

Degree nStepCheck at x

At x = 1: original integrand = 2.718282.

Derivative of displayed known terms = 8.154845; derivative of the remainder = −5.436564; total = 2.718282.

Iₘ means an integral of xᵐeˣ. The remainder is not discarded; its derivative completes the original integrand.

02 / Apply parts twice to a quadratic

The second integral carries the first coefficient.

Find ∫x²eˣ dx.Worked example

I = x²eˣ − 2∫xeˣ dx

First use u = x² and v′ = eˣ.

∫xeˣ dx = xeˣ − eˣ

Use parts again on the remaining product.

I = x²eˣ − 2(xeˣ − eˣ) + C

Keep brackets around the entire second result.

eˣ(x² − 2x + 2) + C

Distribute −2 carefully.

Watch: reduce the degree and keep the remainder

Pause, replay or seek freely. The notes explain the same idea and stay in view.

01 · Complete a first step

After I = ∫x³eˣ dx is integrated by parts once, what is the remainder?

Hint

Differentiate x³.

Worked solution

I = x³eˣ − 3∫x²eˣ dx. The coefficient −3 must remain.

03 / Track factorial-like coefficients

Repeated derivatives create 3, then 6, then 6.

Find ∫x³eˣ dx.Worked example

x³eˣ − 3∫x²eˣ dx

Reduce the polynomial degree.

x³eˣ − 3eˣ(x² − 2x + 2) + C

Use the quadratic result.

eˣ(x³ − 3x² + 6x − 6) + C

The signs alternate for this exponential example.

02 · Check the last sign

What is the constant term inside the polynomial multiplying eˣ in ∫x³eˣ dx?

Hint

Expand −3(x² − 2x + 2).

Worked solution

−6.

03 · Differentiate to check

For P = x³ − 3x² + 6x − 6, find P + P′.

Hint

Collect like powers after differentiating P.

Worked solution

x³. Therefore d(eˣP)/dx = x³eˣ.

04 / Include the exponential rate at every stage

Each integration of e²ˣ contributes a factor 1/2.

Find ∫x²e²ˣ dx.Worked example

(x²/2)e²ˣ − ∫xe²ˣ dx

First use v = e²ˣ/2.

∫xe²ˣ dx = (x/2)e²ˣ − (1/4)e²ˣ

Integrate the remainder by parts.

e²ˣ(x²/2 − x/2 + 1/4) + C

Subtract the entire remainder.

04 · A decaying exponential

Find ∫x²e⁻ˣ dx.

Hint

At each stage, integrating e⁻ˣ adds a minus sign.

Worked solution

−e⁻ˣ(x² + 2x + 2) + C.

05 · A linear combination

Find ∫(x² + 1)eˣ dx.

Hint

Add ∫eˣ dx to the quadratic result.

Worked solution

eˣ(x² − 2x + 3) + C.

05 / Keep the sine-cosine cycle visible

Signs need checking at every integration.

Find ∫x²cos x dx.Worked example

x²sin x − 2∫x sin x dx

First integrate cosine to sine.

∫x sin x dx = −x cos x + sin x

The next primitive changes sign.

x²sin x + 2x cos x − 2sin x + C

Distribute the coefficient −2.

06 · A sine product

Find ∫x²sin x dx.

Hint

The first step is −x²cos x + 2∫x cos x dx.

Worked solution

−x²cos x + 2x sin x + 2cos x + C.

07 · A trig scale factor

Find ∫x²cos(2x) dx.

Hint

First use v = sin(2x)/2.

Worked solution

(x²/2)sin(2x) + (x/2)cos(2x) − (1/4)sin(2x) + C.

06 / Reduce a logarithmic power

Write the integrand as a product with 1.

Find ∫(ln x)² dx for x > 0.Worked example

u = (ln x)²; v′ = 1

Then u′ = 2ln x/x and v = x.

x(ln x)² − 2∫ln x dx

One logarithmic power has disappeared.

x(ln x)² − 2(x ln x − x) + C

Use the known integral of ln x.

x[(ln x)² − 2ln x + 2] + C

Keep the whole expression.

08 · A cubic logarithmic power

State ∫(ln x)³ dx for x > 0.

Hint

The recurrence is x(ln x)³ − 3∫(ln x)² dx.

Worked solution

x[(ln x)³ − 3(ln x)² + 6ln x − 6] + C.

09 · A power times log squared

Find ∫x(ln x)² dx for x > 0.

Hint

Use v = x²/2, leaving ∫x ln x dx.

Worked solution

(x²/2)(ln x)² − (x²/2)ln x + x²/4 + C.

07 / Recognise when the original integral returns

Name it I and solve the resulting equation.

Find I = ∫eˣsin x dx.Worked example

I = eˣsin x − ∫eˣcos x dx

Choose the trig factor as u and eˣ as v′.

J = ∫eˣcos x dx = eˣcos x + I

A second application returns the original integral.

I = eˣsin x − eˣcos x − I

Substitute the whole expression for J.

2I = eˣ(sin x − cos x)

Collect the I terms.

I = (1/2)eˣ(sin x − cos x) + C

Differentiate to verify; one arbitrary constant is retained.

10 · The companion cyclic integral

Find ∫eˣcos x dx.

Hint

Use J = eˣcos x + I from the example.

Worked solution

(1/2)eˣ(sin x + cos x) + C.

08 / Track coefficients when the cycle returns

The coefficient of I need not be one.

Find I = ∫e²ˣsin x dx.Worked example

I = (1/2)e²ˣsin x − (1/2)J

Let J = ∫e²ˣcos x dx.

J = (1/2)e²ˣcos x + (1/2)I

Apply parts again, keeping the scale 1/2.

I = (1/2)e²ˣsin x − (1/4)e²ˣcos x − (1/4)I

Substitute J with brackets.

(5/4)I = (1/4)e²ˣ(2sin x − cos x)

Move the repeated integral to the left.

I = (1/5)e²ˣ(2sin x − cos x) + C

Divide by 5/4.

11 · Solve the integral equation

If I = A − I/4, express I in terms of A.

Hint

Add I/4 to both sides.

Worked solution

I = 4A/5.

12 · A cosine with rate two

Find ∫e²ˣcos x dx.

Hint

Use the companion relation J from the example.

Worked solution

(1/5)e²ˣ(2cos x + sin x) + C.

09 / Evaluate only after the bookkeeping is complete

A checked primitive is often the simplest definite route.

Evaluate ∫₀¹ x²eˣ dx.Worked example

Use F = eˣ(x² − 2x + 2)

This primitive already contains both parts steps.

F(1) = e; F(0) = 2

Do not assume the lower value is zero.

e − 2

Subtract upper minus lower.

Evaluate ∫₁ᵉ (ln x)² dx.Worked example

Use F = x[(ln x)² − 2ln x + 2]

The interval lies inside x > 0.

F(e) = e; F(1) = 2

Use ln e = 1 and ln 1 = 0.

e − 2

The result is positive.

13 · A definite trig polynomial

Evaluate ∫₀^(π/2) x²cos x dx.

Hint

Use x²sin x + 2x cos x − 2sin x.

Worked solution

π²/4 − 2.

10 / Keep the remainder until it is evaluated

A reduction is an identity, not permission to drop a term.

Writing I₂ = x²eˣ − 2I₁ is a useful first step, but x²eˣ is not the answer. The remainder is responsible for cancelling the extra derivative terms. In a cyclic problem, do not keep applying parts forever: solve the equation for the named integral.

Indefinite-integral equations may be handled using consistent primitive representatives during the algebra, with one arbitrary constant added to the final result. Do not introduce several unrelated constants and then equate them without explanation.

14 · Detect a dropped remainder

Why is x²eˣ not a primitive of x²eˣ?

Hint

Apply the product rule.

Worked solution

Its derivative is x²eˣ + 2xeˣ. The missing correction must cancel 2xeˣ.

15 · Detect a cyclic sign error

A student obtains I = A − I and concludes I = 0. Correct the algebra.

Hint

Add I to both sides.

Worked solution

2I = A, so I = A/2. This does not imply I is zero unless A is zero.

11 / Reduce or solve the cycle

Keep coefficients, signs and endpoint values visible.

Differentiate the factor that becomes simpler.
Carry the coefficient into every repeated step.
Retain brackets when substituting a remaining integral.
If the original integral returns, solve for it algebraically.
Differentiate the final expression and then evaluate any bounds.

16 · A final exact value

Evaluate ∫₀^π eˣsin x dx.

Hint

Use (1/2)eˣ(sin x − cos x).

Worked solution

(e^π + 1)/2.

Section 1 of 11 · Keep going while the remainder simplifies