Choosing u dv in Integration by Parts Questions

Choosing u dv under exam pressure

🧭 Why choosing u and dv is the real skill

Most students learn the integration by parts formula early and assume the hard part is memorising it. In exams, the opposite is true. The formula is easy to write down, but the success of the method depends almost entirely on the choice of $u$ and $dv$ . When the choice is poor, the integral becomes harder than the original, or worse, it loops back to itself with no progress.

That is why integration by parts is a favourite examiner topic: it tests decision-making, not just calculus. Under pressure, students tend to choose $u$ and $dv$ based on instinct rather than strategy. This blog builds a reliable routine so you can choose calmly and protect marks. This is one of those A Level Maths topics explained areas where structure beats speed every time.

This decision relies on understanding the integration by parts structure, introduced in Integration by Parts — Method & Exam Insight.

🔙 Previous topic:

The hesitation students feel when choosing $u$ and $d v$ is very similar to what happens in optimisation, so if setting up equations under pressure in optimisation exams still feels uncomfortable, that’s usually where the uncertainty starts.

📘 What examiners are really looking for

Examiners rarely set integration by parts questions to test complicated differentiation or integration. The derivatives are usually straightforward. The integrals are usually basic. What they are testing is whether you can set up the method so that the new integral is simpler than the one you started with. In many mark schemes, the first method mark is awarded for a sensible choice of $u$ and $dv$ and for writing the correct substitution into the formula.

Another method mark often depends on showing clear working for $du$ and $v$ . If your choice causes the integral to get messier, the examiner can see immediately that the method is not secure. That is why choosing well is a scoring skill. It aligns with an A Level Maths revision approach examiners like: reasoned method, not trial and error.

🧠 Integration by parts – the one formula you must use correctly

The standard formula is:
$\int u,dv = uv – \int v,du$

The formula itself is not the difficult part. The difficulty is in ensuring that:

$u$ differentiates to something simpler, and
$dv$ integrates to something manageable,
so that $\int v,du$ is easier than the original integral.

A simple exam check is this: after choosing $u$ and $dv$ , look at what $v,du$ will become. If that product looks worse than what you started with, change your choice immediately. This is how strong students avoid wasting time.

🧮 A practical rule for choosing u

A useful exam heuristic is to choose $u$ as the part that becomes simpler when differentiated. For example:

logarithms: $\ln x$ becomes $\frac{1}{x}$
inverse trig: $\arctan x$ becomes $\frac{1}{1+x^2}$
polynomials: $x^n$ reduces degree each time
exponentials and trig do not become simpler when differentiated, they tend to cycle.

This explains a common exam pattern. When you see a product like $x e^x$ or $x\sin x$ , $u$ is usually the polynomial. That choice forces progress because the polynomial shrinks each differentiation until it disappears. If you choose the exponential or trig as $u$ , nothing gets simpler, and the method becomes fragile.

✏️ A practical rule for choosing dv

Choose $dv$ as the part you can integrate immediately and cleanly. If $dv$ is difficult to integrate, you will get stuck before you even start. In exam conditions, it is almost always safer to let:

$dv = e^x,dx$ , because $v = e^x$
$dv = \sin x,dx$ , because $v = -\cos x$
$dv = \cos x,dx$ , because $v = \sin x$
$dv = \frac{1}{x},dx$ , because $v = \ln x$ (where appropriate)

A common mistake is choosing $dv$ as the “bigger-looking” expression rather than the one that integrates neatly. Examiners set traps where a tempting $dv$ choice leads to messy integration. Your goal is not to be brave. Your goal is to be efficient.

🧠 Why looping happens and how to spot it early

Looping occurs when applying integration by parts produces an integral that is essentially the same as the original. This often happens with products of trig and exponential functions, or repeated cycles such as $\int e^x\sin x,dx$ . Looping is not always a sign of failure — sometimes it is part of the intended method — but you must recognise it and handle it correctly.

The key is to label the original integral as $I$ and be prepared to rearrange at the end. Students lose marks when they loop without noticing and repeat steps aimlessly. Examiners reward students who write $I = \int …$ early and show clean algebraic rearrangement later.

🧪 Complete Deep Exam Question with Full Solution

📄 Exam Question

Evaluate
$\int x^2 e^x , dx$
and explain clearly why your choices of $u$ and $dv$ are sensible.

✅ Full Detailed Solution (with reasoning)

🧠 Step 1: Decide on u and dv (and justify the choice)

We have a product of a polynomial $x^2$ and an exponential $e^x$ .
A strong exam choice is:

$u = x^2$ because differentiating reduces it to $2x$ , then $2$ , then $0$ . This guarantees progress.
$dv = e^x,dx$ because it integrates immediately to $v = e^x$ without creating extra complexity.

So we choose:
$u = x^2 \quad \Rightarrow \quad du = 2x,dx$
$dv = e^x,dx \quad \Rightarrow \quad v = e^x$

This choice is sensible because the new integral $\int v,du$ will involve $2x e^x$ , which is simpler than $x^2 e^x$ due to the lower power of $x$ .

🧮 Step 2: Apply integration by parts once

Use:
$\int u,dv = uv – \int v,du$

So:
$\int x^2 e^x,dx = x^2 e^x – \int e^x(2x),dx$
$= x^2 e^x – 2\int x e^x,dx$

At this stage we have reduced the problem to evaluating:
$\int x e^x,dx$
which is simpler because the polynomial degree has dropped from 2 to 1.

🧠 Step 3: Apply integration by parts a second time (again explain the choice)

To evaluate $\int x e^x,dx$ , we use the same logic:

choose $u = x$ because it differentiates to $1$
choose $dv = e^x,dx$ because it integrates to $e^x$

So:
$u = x \Rightarrow du = dx$
$dv = e^x,dx \Rightarrow v = e^x$

Apply the formula:
$\int x e^x,dx = x e^x – \int e^x , dx$
$= x e^x - e^x$
$= e^x(x - 1)$

🧩 Step 4: Substitute back and simplify fully

We had:
$\int x^2 e^x,dx = x^2 e^x – 2\int x e^x,dx$

Substitute:
$= x^2 e^x – 2\left(e^x(x – 1)\right)$
$= x^2 e^x - 2xe^x + 2e^x$

Factor out $e^x$ :
$= e^x(x^2 - 2x + 2) + C$

✅ Final Answer

$\boxed{\int x^2 e^x,dx = e^x(x^2 – 2x + 2) + C}$

🧠 Examiner insight: what earns full marks here

Examiners reward students who show a clear reduction in complexity. The key method marks are usually earned for:

a sensible first choice of $u$ and $dv$ ,
correct identification of $du$ and $v$ ,
correct application of the formula,
and correct completion of the second integration by parts.

Students often lose marks by skipping the second by-parts step and trying to integrate $\int x e^x,dx$ by guessing. That may work sometimes, but it is not reliable and it hides method. Another common loss is failing to simplify at the end, leaving a correct but messy expression. Examiners prefer clean final forms because it shows control.

🎯 Final exam takeaway

Integration by parts is not about memorising a formula. It is about making choices that simplify the integral step by step. In exams, choose $u$ to be the part that becomes simpler when differentiated, and choose $dv$ to be the part that integrates cleanly. Always check that the new integral is simpler than the original before committing. With consistent practice — supported by a A Level Maths Revision Course for every exam board — integration by parts becomes predictable and scoreable rather than stressful.

✍️ Author Bio

👨‍🏫 S. Mahandru

When students struggle with integration by parts, it is rarely because the calculus is too hard. It is because the first choice is rushed. Teaching focuses on building a simple decision routine so method marks are protected under exam pressure.

🧭 Next topic:

Once you’re confident about choosing $u$ and $d v$ , the next thing that tends to trip students up is common sign errors that lose marks in integration by parts, especially when negatives start stacking up mid-solution.

❓ FAQs

🧭 How do I choose u and dv quickly in exams without guessing?

Start by identifying which part becomes simpler when differentiated. Polynomials reduce degree each time, and $\ln x$ becomes $\frac{1}{x}$ , which is usually simpler. Then look for the part that integrates cleanly, such as $e^x$ , $\sin x$ , or $\cos x$ . Your goal is to make $\int v,du$ simpler than the original integral.

If your choice makes the new integral more complicated, change immediately. A fast check is to mentally form $v,du$ and see if it looks worse. With practice, this becomes a routine rather than a decision. Examiners reward sensible choices even if later arithmetic slips. The more you show structure, the more method marks you protect.

🧠 What if both choices seem to work?

If both seem workable, choose the one that reduces complexity more reliably. Polynomials are usually the safest $u$ because they eventually become zero. If no polynomial exists, $\ln x$ is often the next safest choice because it simplifies dramatically when differentiated. Avoid choosing trig or exponential as $u$ unless the other part does not simplify.

Another clue is whether $dv$ integrates immediately without producing awkward fractions or extra functions. In exams, “clean integration” is a major advantage. If you are unsure, write both options briefly and compare the resulting $\int v,du$ . Choose the one with simpler algebra. This is faster than committing to a poor method and restarting later.

⚖️ Why do some integration by parts questions “loop” back to the same integral?

Looping happens when the new integral after by-parts is essentially the same as the original. This often occurs with products like $e^x\sin x$ or $e^x\cos x$ , where differentiating and integrating cycles. Examiners sometimes intend this and expect students to label the original integral as $I$ . After two applications, $I$ reappears on the right-hand side.

The correct move is to collect the $I$ terms and rearrange algebraically. Students lose marks when they repeat steps without recognising the loop. The loop is not a dead end; it is a prompt to rearrange. Writing $I = \int …$ at the start signals to the examiner that you understand the strategy. Once you’ve rearranged, the remaining work is straightforward. This is one of the clearest ways examiners test method under pressure.