9709 P3 §3.5 — Integration by Parts I

💡 Lead In: The Product That Won't Split

The Scenario

Last week we integrated functions like $e^{2x}$, $\sin(3x)$, and $\dfrac{1}{2x-1}$ by reversing the chain rule. But now consider the integral

$$\int x \sin 2x \, \text{d}x$$

This is a product of two functions: $x$ (algebraic) and $\sin 2x$ (trigonometric). A student tries to “split” the integral:

“$\displaystyle\int x \sin 2x \, \text{d}x = \left(\int x \, \text{d}x\right) \cdot \left(\int \sin 2x \, \text{d}x\right) = \dfrac{x^2}{2} \cdot \left(-\dfrac{1}{2}\cos 2x\right)$”

But checking by differentiation shows this is wrong. Integration does not split over products. None of our W6 techniques handle a product directly. We need a new technique — one that reverses the product rule, just as W6 reversed the chain rule.

The Question

We know the product rule for differentiation:

$$\frac{\text{d}}{\text{d}x}(uv) = u'v + uv'$$

Can you integrate both sides of this identity and rearrange to find a formula for $\int u \, v' \, \text{d}x$?

What new formula do you get? And how could you use it to evaluate $\int x \sin 2x \, \text{d}x$?

💡 The Bridge (click to reveal)

Integrate both sides of $(uv)' = u'v + uv'$:

$$uv = \int u'v \, \text{d}x + \int u v' \, \text{d}x$$

Rearranging gives the integration by parts formula:

$$\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$$

where you differentiate $u$ to get $u'$, and integrate $v'$ to get $v$.

For $\int x \sin 2x \, \text{d}x$: choose $u = x$ (so $u' = 1$, simpler) and $v' = \sin 2x$ (so $v = -\frac{1}{2}\cos 2x$). Then $\int x \sin 2x \, \text{d}x = x \cdot \left(-\frac{1}{2}\cos 2x\right) - \int \left(-\frac{1}{2}\cos 2x\right) \text{d}x$. Today we learn this technique in full.

📖 Key Concepts: Integration by Parts

The product rule says $\dfrac{\text{d}}{\text{d}x}(uv) = u'v + uv'$. Integrating both sides and rearranging gives the integration by parts formula — the technique for integrating a product of two functions.

Key relation 1 — The integration by parts formula:

If $u = u(x)$ and $v = v(x)$ are differentiable functions, then:

$$\boxed{\;\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x\;}$$

The key idea: differentiate one factor to get $u'$, integrate the other to get $v$, multiply $u$ by $v$, then subtract the integral of $u'v$.

Key relation 2 — The LIATE rule for choosing $u$:

A by-parts integral such as $\int x \sin 2x \, \text{d}x$ has two factors: one will be differentiated (written $u$) and one will be integrated (written $v'$). The method only helps if the new integral $\int u' v \, \text{d}x$ is simpler than the original, so the split must be chosen deliberately.

LIATE is an acronym for the five function types, listed from the best choice of $u$ (left) to the worst (right):

Logarithmic

$\ln x$, $\log_2 x$

Inverse trig

$\sin^{-1} x$, $\tan^{-1} x$

Algebraic

$x$, $x^2$, $3x+1$

Trigonometric

$\sin x$, $\cos 2x$

Exponential

$e^x$, $e^{2x}$

← earlier in LIATE: choose as $u$ (differentiate) | later in LIATE: choose as $v'$ (integrate) →

Reading the rule: compare the two factors; the one whose type comes earlier in LIATE becomes $u$, and the other becomes $v'$.

Why this order? Differentiating L, I or A gives something simpler — $\ln x \to \dfrac{1}{x}$, $\tan^{-1} x \to \dfrac{1}{1+x^2}$, and $x^2 \to 2x \to 2$. Differentiating T or E simplifies nothing ($\sin x \to \cos x$, $e^x \to e^x$), but both are easy to integrate. So $u$ is taken from the left of the list and $v'$ from the right.

Apply it — $\int x \ln x \, \text{d}x$:

$\ln x$ is Logarithmic (L) and $x$ is Algebraic (A). L comes before A, so $u = \ln x$ and $v' = x$. Then $u' = \dfrac{1}{x}$ and $v = \dfrac{x^2}{2}$, and the residual $\dfrac{1}{2}\displaystyle\int x \, \text{d}x$ is a basic power integral. The reverse choice ($u = x$, $v' = \ln x$) fails at once, because $\ln x$ cannot be integrated directly.

Key relation 3 — Repeated integration by parts:

Sometimes the integral $\int u' v \, \text{d}x$ that results from the first application is still a product. In that case, apply integration by parts again. Each application of LIATE reduces the algebraic power (e.g. $x^2 \to 2x \to 2$) until the remaining integral is directly integrable.

Example: $\displaystyle\int x^2 e^{-x} \, \text{d}x$ requires two applications — first $u = x^2$ (so $u' = 2x$), then $u = 2x$ (so $u' = 2$, which integrates directly).

Summary — the by-parts strategy:

Identify the two factors in the integrand.
Use LIATE to choose $u$ (the factor to differentiate) and $v'$ (the factor to integrate).
Compute $u'$ and $v = \int v' \, \text{d}x$.
Apply: $\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$.
If $\int u' v \, \text{d}x$ is still a product, apply by parts again.
Always verify by differentiating your answer.

Group Activity 1: Reveal the By-Parts Method — Approve Every Step

In groups of three you must approve each line before it is shown. At every stage, pick the line you believe comes next in evaluating $\displaystyle\int x e^{2x}\, \text{d}x$. A correct pick reveals the line and unlocks the next stage; a wrong pick turns red and explains the trap. Only four wrong picks in the whole run earns you the top band. Nobody clicks until all three members agree on the line.

Group rule: say the reasoning out loud before anyone clicks — which factor is $u$, which is $v'$, and why. If you disagree, justify your line; that is where the learning is.

Individual Activity 2: By-Parts Function Machine — Feed the Right Factor

On your own. The machine turns one chosen factor of the integrand into $v$ and returns a residual integral. Use the slider to feed a different factor as $u$ and watch the residual bar: the dashed line is the simplest residual the machine can produce. Predict which factor is best before you move the slider, then press Check split. Match the dashed target on all three missions and answer the two follow-ups.

Feed as $u$:

The factor you did not feed as $u$ becomes $v'$; the machine integrates it and subtracts.

Follow-up 1 — In LIATE, which type is chosen as $u$ first?

Follow-up 2 — Why does the LIATE choice shrink the residual?

Dashed line = the simplest residual (the target). Emerald bar = the residual your chosen factor produces. A lower bar is a better split.

Group · Guided Worked Example 1: $\int x e^x \, \text{d}x$ by Parts Synthesis · simplest A × E case

Find $\displaystyle\int x e^x \, \text{d}x$.

Agree in your group which factor must be $u$ and which must be $v'$, then check your reasoning against the solution.

👁 Reveal Solution

Step 1: Choose $u$ and $v'$ using LIATE

The factors are $x$ (Algebraic) and $e^x$ (Exponential). A comes before E in LIATE, so the algebraic factor is the one to differentiate:

$u = x$, $v' = e^x$.

The reverse choice would give $v = \dfrac{x^2}{2}$ and $u' = e^x$, leaving $\displaystyle\int \dfrac{x^2}{2}e^x \, \text{d}x$ — worse than the original integral.

Step 2: Compute $u'$ and $v$

$u' = 1$, $v = \displaystyle\int e^x \, \text{d}x = e^x$.

Step 3: Apply the by-parts formula

$\displaystyle\int x e^x \, \text{d}x = uv - \int u' v \, \text{d}x = x e^x - \int e^x \cdot 1 \, \text{d}x = x e^x - e^x + C$.

Step 4: Factorise

$\displaystyle\int x e^x \, \text{d}x = (x - 1)e^x + C$.

Final Answer

$\boxed{\displaystyle\int x e^x \, \text{d}x = (x - 1)e^x + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[(x-1)e^x\right] = 1 \cdot e^x + (x-1) \cdot e^x = e^x + x e^x - e^x = x e^x$. ✓

Individual · Guided Worked Example 2: $\int x^2 \ln x \, \text{d}x$ by Parts Synthesis · L before A

Find $\displaystyle\int x^2 \ln x \, \text{d}x$.

Here the logarithm must be $u$ — decide why before you read the solution.

👁 Reveal Solution

Step 1: Choose $u$ and $v'$ using LIATE

$\ln x$ is Logarithmic (L) and $x^2$ is Algebraic (A). L comes before A, so $u = \ln x$ and $v' = x^2$.

The reverse choice cannot even be started: $v' = \ln x$ cannot be integrated without already using by parts.

Step 2: Compute $u'$ and $v$

$u' = \dfrac{1}{x}$, $v = \displaystyle\int x^2 \, \text{d}x = \dfrac{x^3}{3}$.

Step 3: Apply the by-parts formula

$\displaystyle\int x^2 \ln x \, \text{d}x = uv - \int u' v \, \text{d}x = \ln x \cdot \dfrac{x^3}{3} - \int \dfrac{1}{x} \cdot \dfrac{x^3}{3} \, \text{d}x$

$= \dfrac{x^3}{3}\ln x - \dfrac{1}{3}\displaystyle\int x^2 \, \text{d}x$.

Step 4: Evaluate the remaining integral

$\dfrac{1}{3}\displaystyle\int x^2 \, \text{d}x = \dfrac{1}{3} \cdot \dfrac{x^3}{3} = \dfrac{x^3}{9}$, so $\displaystyle\int x^2 \ln x \, \text{d}x = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$.

Final Answer

$\boxed{\displaystyle\int x^2 \ln x \, \text{d}x = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}\!\left(\dfrac{x^3}{3}\ln x - \dfrac{x^3}{9}\right) = x^2\ln x + \dfrac{x^3}{3} \cdot \dfrac{1}{x} - \dfrac{x^2}{3} = x^2\ln x + \dfrac{x^2}{3} - \dfrac{x^2}{3} = x^2\ln x$. ✓

✏️ Concept Quiz

Score: 0 / 4

1. The integration by parts formula is:

$\int u v' \, \text{d}x = uv + \int u' v \, \text{d}x$

$\int u v' \, \text{d}x = \dfrac{u}{v} - \int u' v \, \text{d}x$

$\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$

$\int u v' \, \text{d}x = uv - \int u v' \, \text{d}x$

Integrating the product rule $(uv)' = u'v + uv'$ gives $uv = \int u'v \, \text{d}x + \int uv' \, \text{d}x$. Rearranging: $\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$. The trap $uv + \int u' v \, \text{d}x$ has the wrong sign. The trap $\int u v' \, \text{d}x = uv - \int u v' \, \text{d}x$ is circular (the same integral appears on both sides).

2. According to the LIATE rule, which factor should be chosen as $u$ in $\int x^2 e^x \, \text{d}x$?

$x^2$ (Algebraic)

$e^x$ (Exponential)

$x^2 e^x$ (the whole product)

$1$ (a constant)

LIATE: L > I > A > T > E. The factor $x^2$ is Algebraic (A) and $e^x$ is Exponential (E). Since A comes before E, choose $u = x^2$ (so $u' = 2x$, simpler) and $v' = e^x$ (so $v = e^x$). The trap $u = e^x$ gives $u' = e^x$ and $v = \frac{x^3}{3}$, making $\int u' v \, \text{d}x = \int \frac{x^3}{3} e^x \, \text{d}x$ — harder than the original.

3. To evaluate $\int \ln x \, \text{d}x$ using integration by parts, the correct choice is:

$u = 1, \;\; v' = \ln x$

$u = \ln x, \;\; v' = 1$

$u = \ln x, \;\; v' = x$

$u = x, \;\; v' = \ln x$

Write $\int \ln x \, \text{d}x = \int (\ln x) \cdot 1 \, \text{d}x$. By LIATE, $\ln x$ (Logarithmic, first priority) is $u$, and $1$ is $v'$ (so $v = x$). The trap $u = 1, v' = \ln x$ is circular — we cannot find $v$ without already knowing $\int \ln x \, \text{d}x$. The choice $v' = x$ would correspond to the integrand $\ln x \cdot x = x \ln x$, not $\ln x$.

4. To evaluate $\int x^2 \sin x \, \text{d}x$, how many times must you apply integration by parts?

Once

Twice

Three times

It cannot be done by parts

First application: $u = x^2$ (so $u' = 2x$), $v' = \sin x$ (so $v = -\cos x$). The resulting integral $\int 2x \cos x \, \text{d}x$ is still a product, so apply by parts a second time: $u = 2x$ (so $u' = 2$), $v' = \cos x$ (so $v = \sin x$). Now $\int 2 \sin x \, \text{d}x = -2 \cos x$ is a basic integral. So two applications are needed. The algebraic power $x^2 \to 2x \to 2$ decreases each time until it becomes a constant.

🎉

You scored 0 / 4

📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Q1 · Synthesis Standard

Find $\displaystyle\int x \sin x \, \text{d}x$.

💡 Hint

Use LIATE: $x$ is Algebraic, $\sin x$ is Trigonometric. Choose $u = x$, $v' = \sin x$.

Then $u' = 1$ and $v = -\cos x$. Apply the by-parts formula.

👁 Reveal Solution

Step 1: Choose $u$ and $v'$

$u = x$ (Algebraic), $v' = \sin x$ (Trigonometric).

Step 2: Compute $u'$ and $v$

$u' = 1$, $v = \displaystyle\int \sin x \, \text{d}x = -\cos x$.

Step 3: Apply the formula

$\displaystyle\int x \sin x \, \text{d}x = uv - \int u' v \, \text{d}x = x(-\cos x) - \int (-\cos x) \cdot 1 \, \text{d}x = -x \cos x + \int \cos x \, \text{d}x$.

$= -x \cos x + \sin x + C$.

Final Answer

$\boxed{-x \cos x + \sin x + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}(-x \cos x + \sin x) = -\cos x + x \sin x + \cos x = x \sin x$. ✓

Q2 · Synthesis Medium

Find $\displaystyle\int x^2 \cos x \, \text{d}x$.

(This requires applying integration by parts twice.)

💡 Hint 1

Choose $u = x^2$ (Algebraic), $v' = \cos x$ (Trigonometric).

Then $u' = 2x$ and $v = \sin x$. Apply the formula once.

💡 Hint 2

After the first application, you get $\int x^2 \cos x \, \text{d}x = x^2 \sin x - 2\int x \sin x \, \text{d}x$.

The remaining integral $\int x \sin x \, \text{d}x$ is still a product — apply by parts again with $u = x$, $v' = \sin x$.

👁 Reveal Solution

Step 1: First by-parts application

$u = x^2$, $v' = \cos x$ $\Rightarrow$ $u' = 2x$, $v = \sin x$.

$\displaystyle\int x^2 \cos x \, \text{d}x = x^2 \sin x - \int u' v \, \text{d}x = x^2 \sin x - \int 2x \sin x \, \text{d}x = x^2 \sin x - 2\int x \sin x \, \text{d}x$.

Step 2: Second by-parts application on $\int x \sin x \, \text{d}x$

$u = x$, $v' = \sin x$ $\Rightarrow$ $u' = 1$, $v = -\cos x$.

$\displaystyle\int x \sin x \, \text{d}x = -x \cos x - \int (-\cos x) \cdot 1 \, \text{d}x = -x \cos x + \sin x$.

Step 3: Combine

$\displaystyle\int x^2 \cos x \, \text{d}x = x^2 \sin x - 2(-x \cos x + \sin x) + C$.

$= x^2 \sin x + 2x \cos x - 2 \sin x + C$.

Final Answer

$\boxed{x^2 \sin x + 2x \cos x - 2 \sin x + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}(x^2 \sin x + 2x \cos x - 2 \sin x) = 2x \sin x + x^2 \cos x + 2\cos x - 2x\sin x - 2\cos x = x^2 \cos x$. ✓

Q3 · Synthesis Medium

Find $\displaystyle\int x e^{2x} \, \text{d}x$.

💡 Hint

Use LIATE: $u = x$ (Algebraic), $v' = e^{2x}$ (Exponential).

Then $u' = 1$ and $v = \dfrac{1}{2} e^{2x}$ (reverse chain rule from W6). Apply the formula.

👁 Reveal Solution

Step 1: Choose $u$ and $v'$

$u = x$ (Algebraic), $v' = e^{2x}$ (Exponential).

Step 2: Compute $u'$ and $v$

$u' = 1$, $v = \displaystyle\int e^{2x} \, \text{d}x = \dfrac{1}{2} e^{2x}$.

Step 3: Apply the formula

$\displaystyle\int x e^{2x} \, \text{d}x = uv - \int u' v \, \text{d}x = x \cdot \dfrac{1}{2} e^{2x} - \int \dfrac{1}{2} e^{2x} \cdot 1 \, \text{d}x$.

$= \dfrac{x}{2} e^{2x} - \dfrac{1}{2} \cdot \dfrac{1}{2} e^{2x} + C = \dfrac{x}{2} e^{2x} - \dfrac{1}{4} e^{2x} + C$.

Step 4: Factorise

$= \dfrac{1}{4}(2x - 1) e^{2x} + C$.

Final Answer

$\boxed{\dfrac{1}{4}(2x - 1) e^{2x} + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[\dfrac{1}{4}(2x-1)e^{2x}\right] = \dfrac{1}{4}\left(2 \cdot e^{2x} + (2x-1) \cdot 2e^{2x}\right) = \dfrac{1}{4}\left(2 + 4x - 2\right)e^{2x} = \dfrac{1}{4} \cdot 4x \cdot e^{2x} = x e^{2x}$. ✓

Q4 · Synthesis Challenging

Find $\displaystyle\int x^2 e^{-x} \, \text{d}x$.

(Two applications of by parts are needed. Choose $u$ from the algebraic factor each time.)

💡 Hint 1

First application: $u = x^2$ (Algebraic), $v' = e^{-x}$ (Exponential), so $v = -e^{-x}$.

💡 Hint 2

The remaining integral is $2\displaystyle\int x e^{-x} \, \text{d}x$ — still a product, so apply by parts again with $u = x$, $v' = e^{-x}$.

👁 Reveal Solution

Step 1: First application — choose $u$ and $v'$

$u = x^2$ (Algebraic), $v' = e^{-x}$ (Exponential) ⇒ $u' = 2x$, $v = -e^{-x}$.

$\displaystyle\int x^2 e^{-x} \, \text{d}x = -x^2 e^{-x} - \int (-e^{-x})(2x) \, \text{d}x = -x^2 e^{-x} + 2\int x e^{-x} \, \text{d}x$.

Step 2: Second application on $\int x e^{-x} \, \text{d}x$

$u = x$, $v' = e^{-x}$ ⇒ $u' = 1$, $v = -e^{-x}$.

$\displaystyle\int x e^{-x} \, \text{d}x = -x e^{-x} - \int (-e^{-x}) \cdot 1 \, \text{d}x = -x e^{-x} - e^{-x}$.

Step 3: Combine

$\displaystyle\int x^2 e^{-x} \, \text{d}x = -x^2 e^{-x} + 2\left(-x e^{-x} - e^{-x}\right) + C = -\left(x^2 + 2x + 2\right)e^{-x} + C$.

Final Answer

$\boxed{\displaystyle\int x^2 e^{-x} \, \text{d}x = -\left(x^2 + 2x + 2\right)e^{-x} + C}$

Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[-\left(x^2+2x+2\right)e^{-x}\right] = -(2x+2)e^{-x} + \left(x^2+2x+2\right)e^{-x} = x^2 e^{-x}$. ✓

🎫

Exit Ticket

Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.

I can state the integration by parts formula $\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$ and derive it by integrating the product rule. I can use the LIATE rule (Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential) to choose $u$ and $v'$ correctly for any product of two functions. I can evaluate integrals like $\int x \sin 2x \, \text{d}x$, $\int x e^x \, \text{d}x$, and $\int \ln x \, \text{d}x$ using a single application of integration by parts. I can apply integration by parts repeatedly (e.g. $\int x^2 e^{-x} \, \text{d}x$, $\int x^2 \cos x \, \text{d}x$) and verify my answer by differentiation.