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Integration by parts

Learn integration by parts from the product rule. Choose u and dv, integrate polynomial-exponential and trig products, handle ln x and evaluate definite integrals with original worked practice.

Before you startThe product rule, standard integrals and definite-integral notation.

01 / Reverse the product rule

A product primitive needs a correction term.

∫u v′ dx = uv − ∫v u′ dx

Also written ∫u dv = uv − ∫v du.

Differentiating uv produces u′v as well as uv′. If uv′ is the integrand you want, subtract the integral of the extra term u′v. Parts is useful when that remaining integral is simpler.

The existing integration by parts interactive lesson is still available as a companion.

Subtract the extra product-rule termExplore

ExampleStepCheck at x

At x = 1: (uv)′ = 5.436564; u′v = 2.718282; uv′ = 2.718282.

Product derivative − extra term = 2.718282, matching the original integrand.

Start with the product rule, which creates two terms.

The numerical values illustrate the identity. The formula works throughout the stated domain; the logarithmic example uses x > 0.

02 / Derive the formula in three lines

The minus sign comes from rearranging the product rule.

Start with the product derivative.Worked example

(uv)′ = u′v + uv′

There are two contributions.

uv′ = (uv)′ − u′v

Isolate the wanted term.

∫uv′ dx = uv − ∫u′v dx

Integrate both sides. Include one arbitrary constant when evaluating the final primitive.

Watch: remove the extra product-rule contribution

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

01 · Name the correction

For ∫xeˣ dx with u = x and v′ = eˣ, what is u′v?

Hint

Differentiate x and integrate eˣ.

Worked solution

u′v = eˣ. Its integral is the correction to subtract from xeˣ.

03 / Differentiate one factor and integrate the other

Make the remaining integral easier.

Find ∫xeˣ dx.Worked example

u = x; v′ = eˣ

Differentiating x lowers its degree.

u′ = 1; v = eˣ

Write this short table before substituting.

xeˣ − ∫eˣ dx

Use uv minus the remaining integral.

xeˣ − eˣ + C

Differentiate: eˣ + xeˣ − eˣ = xeˣ.

Choosing u = eˣ and v′ = x is valid algebraically, but leaves an integral involving x²eˣ, which is harder. A useful choice simplifies the remainder.

02 · Choose the useful u

For ∫x cos x dx, which factor is a useful choice for u?

Hint

Prefer the factor that simplifies when differentiated.

Worked solution

u = x, with v′ = cos x, so u′ = 1 and v = sin x.

03 · Do not differentiate both factors

If v′ = e²ˣ, what is v?

Hint

You need an antiderivative, not a derivative.

Worked solution

v = (1/2)e²ˣ.

04 / Keep the exponential scale factor

The factor in v appears in both terms of the formula.

Find ∫xe²ˣ dx.Worked example

u = x; u′ = 1; v′ = e²ˣ; v = e²ˣ/2

Integrate e²ˣ with its factor 1/2.

(x/2)e²ˣ − (1/2)∫e²ˣ dx

The same v belongs in uv and the remaining integral.

(x/2)e²ˣ − (1/4)e²ˣ + C

The second integration supplies another factor 1/2.

04 · A negative exponential rate

Find ∫xe⁻ˣ dx.

Hint

v = −e⁻ˣ.

Worked solution

−xe⁻ˣ − e⁻ˣ + C = −(x + 1)e⁻ˣ + C.

05 · A linear factor

Find ∫(3x + 2)eˣ dx.

Hint

Use u = 3x + 2, so u′ = 3.

Worked solution

(3x + 2)eˣ − 3eˣ + C = (3x − 1)eˣ + C.

05 / Track the trig signs

A negative primitive changes the correction term.

Find ∫x cos x dx.Worked example

u = x; v′ = cos x; v = sin x

The derivative of u is 1.

x sin x − ∫sin x dx

Substitute into the parts formula.

x sin x + cos x + C

The integral of sin x is −cos x.

Find ∫x sin x dx.Worked example

u = x; v = −cos x

Integrating sine creates a minus sign.

−x cos x + ∫cos x dx

Subtracting the negative v changes the sign.

−x cos x + sin x + C

Differentiate to check.

06 · A scaled cosine

Find ∫x cos(2x) dx.

Hint

v = sin(2x)/2.

Worked solution

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

07 · A shifted polynomial factor

Find ∫(x + 2)sin x dx.

Hint

Use u = x + 2 and v = −cos x.

Worked solution

−(x + 2)cos x + sin x + C.

06 / Write ln x as a product with 1

Its derivative simplifies even though its primitive is initially unknown.

Find ∫ln x dx for x > 0.Worked example

u = ln x; v′ = 1

The hidden second factor is 1.

u′ = 1/x; v = x

Both operations are straightforward.

x ln x − ∫x(1/x) dx

The remaining integrand simplifies to 1.

x ln x − x + C

Differentiate: ln x + 1 − 1 = ln x.

08 · A constant logarithm factor

Find ∫5ln x dx for x > 0.

Hint

Keep the factor 5 outside.

Worked solution

5x ln x − 5x + C.

09 · An absolute-value logarithm

Find ∫ln|x| dx on either interval x < 0 or x > 0.

Hint

The derivative of ln|x| is 1/x on both intervals.

Worked solution

x ln|x| − x + C, with a separate constant permitted on each interval.

07 / Combine a power and a logarithm

Choose the logarithm as the factor to differentiate.

Find ∫x ln x dx for x > 0.Worked example

u = ln x; v′ = x

Differentiating ln x turns it into a reciprocal.

u′ = 1/x; v = x²/2

The remaining product becomes x/2.

(x²/2)ln x − (1/2)∫x dx

Apply parts once.

(x²/2)ln x − x²/4 + C

Check by differentiating the whole expression.

10 · A higher power with one logarithm

Find ∫x²ln x dx for x > 0.

Hint

Use v = x³/3.

Worked solution

(x³/3)ln x − x³/9 + C.

08 / Use the definite parts formula

The boundary term and remaining integral share the original bounds.

∫ₐᵇ u v′ dx = [uv]ₐᵇ − ∫ₐᵇ v u′ dx

Evaluate ∫₀¹ xeˣ dx.Worked example

[xeˣ]₀¹ − ∫₀¹ eˣ dx

Apply the bounds to both pieces.

e − (e − 1)

The remaining integral is upper minus lower.

Keep brackets until the subtraction is complete.

11 · A definite trig product

Evaluate ∫₀^(π/2) x cos x dx.

Hint

Use x sin x + cos x.

Worked solution

π/2 − 1.

12 · Definite logarithm

Evaluate ∫₁ᵉ ln x dx.

Hint

Use x ln x − x.

Worked solution

0 − (−1) = 1.

09 / Subtract the entire remaining integral

A sign error often enters when brackets disappear too early.

Evaluate ∫₀^(π/2) x sin x dx.Worked example

[−x cos x]₀^(π/2) + ∫₀^(π/2) cos x dx

The v = −cos x sign has been carried through.

0 + [sin x]₀^(π/2)

The boundary term is zero at both endpoints.

The positive value agrees with the positive integrand.

13 · Keep endpoint brackets

For ∫₀¹ xe²ˣ dx, evaluate [(x/2)e²ˣ − e²ˣ/4]₀¹.

Hint

The lower endpoint contributes −1/4.

Worked solution

e²/4 − (−1/4) = (e² + 1)/4.

10 / Use differentiation to test the answer

The extra product-rule terms must cancel.

If the derivative still contains an unwanted term, the correction is missing or has the wrong sign. For example, xeˣ alone differentiates to xeˣ + eˣ; subtracting eˣ is essential.

If one application leaves a similar but simpler product, apply parts again. The next lesson develops repeated parts, logarithmic powers and cases where the original integral returns.

14 · Repair the correction sign

A student gives ∫x cos x dx = x sin x − cos x + C. Correct it.

Hint

Differentiate: the two sine terms would add.

Worked solution

x sin x + cos x + C. Its derivative is sin x + x cos x − sin x.

15 · Is one application enough?

With u = x² and v′ = eˣ, what remains after one application of parts?

Hint

u′ = 2x.

Worked solution

x²eˣ − 2∫xeˣ dx. One further application evaluates the remaining integral.

11 / Choose, tabulate, subtract and check

Integrate v′ and differentiate u.

Choose u so that u′ simplifies the remaining integral.
Find v accurately, including signs and scale factors.
Use uv − ∫v u′ dx.
Keep bounds and subtraction brackets for definite integrals.
Differentiate the completed primitive to check cancellation.

16 · A new exponential product

Find ∫(2x − 1)e³ˣ dx.

Hint

Use u = 2x − 1 and v = e³ˣ/3.

Worked solution

[(2x − 1)/3 − 2/9]e³ˣ + C = (6x − 5)e³ˣ/9 + C.

Section 1 of 11 · Reverse the product rule