Mathematics 9709 · Pure Mathematics 3

Integration by parts: integrating a product

By parts is the product rule run backwards.
A product of two unrelated functions ( $x\,e^{x}$ , $x\ln x$ , $x\tan^{-1}x$ ) could only have come from differentiating a product, so this is the tool to use.
The most important choice is which factor you call $u$ .

The formula and the choice of u

$\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$ (textbook 8.6, Key Point 8.4)

You choose $u$ (the part you differentiate) and $\dfrac{dv}{dx}$ (the part you integrate).
Pick $u$ to be the part that gets simpler when you differentiate it, and that you could not easily integrate anyway.

A rule you can use without thinking: if $\ln x$ or $\tan^{-1}x$ is there, that is $u$ (you cannot integrate them nicely). Otherwise $u$ = the power of $x$ .

Common mistake

The wrong choice of

u

wastes the whole question. You cannot integrate

\ln x

to get

v

, so for

x\ln x

you must take

u=\ln x

. If your leftover integral comes out harder than the one you started with, that is the sign to swap.

Worked example

9709/31 Nov 2024 Q2: $\int_1^3 x^2\ln 3x\,dx$

$\ln 3x$ is present → take $u=\ln 3x$ , $\dfrac{dv}{dx}=x^2$ .
Then $\dfrac{du}{dx}=\dfrac{1}{x}$ and $v=\dfrac{x^3}{3}$ .
Apply: $\left[\dfrac{x^3}{3}\ln 3x\right]_1^3 - \int_1^3 \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx$ ; the leftover $\int \dfrac{x^2}{3}\,dx$ is now trivial.

(9709/31 Nov 2024 Q2) wants the answer “in the form $a\ln b + c$ .” Work out the $[uv]$ bracket at both limits and show it clearly; that bracket is worth a method mark. The full value is $\tfrac{53}{3}\ln 3 - \tfrac{26}{9}$ .

Worked example

9709/33 Jun 2020 Q2: $\int_0^1 (2-x)e^{-2x}\,dx$ : the other u

Now there is no $\ln$ or $\tan^{-1}$ , so the rule says $u$ = the polynomial, and the exponential is the part you integrate.

Take $u=2-x$ , $\dfrac{dv}{dx}=e^{-2x}$ . Then $\dfrac{du}{dx}=-1$ and $v=-\tfrac12 e^{-2x}$ .
Apply the formula: $\left[-\tfrac12(2-x)e^{-2x}\right]_0^1 - \int_0^1 \left(-\tfrac12 e^{-2x}\right)(-1)\,dx$ .
The leftover $-\tfrac12\int_0^1 e^{-2x}\,dx=\left[\tfrac14 e^{-2x}\right]_0^1$ is now a one-line integral. Combine the two brackets.

Answer

\int_0^1 (2-x)e^{-2x}\,dx = \tfrac34 - \tfrac14 e^{-2}

(9709/33 Jun 2020 Q2). The choice is the whole question: with no log here, the $x$ -polynomial is the part that gets simpler when you differentiate, so it is $u$ .

Why “differentiate the part that gets simpler” is the right rule

By parts swaps your integral $\int u\,\tfrac{dv}{dx}$ for a new one $\int v\,\tfrac{du}{dx}$ . That swap only helps if the new integral is easier. Differentiating a power of $x$ drops its degree ( $x^2\to 2x\to 2$ ), heading towards a constant. Differentiating $\ln x$ or $\tan^{-1}x$ turns them into algebraic fractions you can actually integrate. So you pick $u$ to be whatever shrinks when you differentiate it. That is the part you want gone from the next integral.

∫ ln x dx: write it as 1 · ln x

Integrating $\ln x$ on its own looks impossible, until you see it is really a product.

Write $\ln x = 1\cdot\ln x$ , then take $u=\ln x$ and $\dfrac{dv}{dx}=1$ (so $v=x$ ):

$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + c$

(textbook Worked Example 8.17). The same $1\cdot(\dots)$ move handles $\int \tan^{-1}x\,dx$ .

Your turn— tap to reveal the worked answer (9709/31 Nov 2025 Q1)

“Find the exact value of $\displaystyle\int_1^2 \ln 3x\,dx$ in the form $a+\ln b$ .” Write it as $1\cdot\ln 3x$ , take $u=\ln 3x$ ( $\tfrac{du}{dx}=\tfrac1x$ ) and $v=x$ : $[x\ln 3x]_1^2-\int_1^2 1\,dx$ . That gives $2\ln 6-\ln 3-1$ , and the log laws fold it down to:

Answer

\int_1^2 \ln 3x\,dx = \ln 12 - 1

(9709/31 Nov 2025 Q1). Same $1\cdot(\dots)$ move as $\int\ln x\,dx$ . The only extra step is folding $2\ln 6-\ln 3$ with log laws.

Applying it twice

When $u$ is a higher power of $x$ (like $x^2\sin x$ or $x^2 e^{x}$ ), one pass turns $x^2$ into $2x$ but still leaves a product behind.
Do by parts a second time on that leftover. The power drops to a constant and it finishes.

Common mistake

On the second pass, the whole job is the signs. The formula's

-\int v\,\tfrac{du}{dx}

term flips the sign each time. Keep track of the minus. Most lost marks here are one dropped sign, not a wrong method. (textbook Worked Example 8.18)

∫ x tan⁻¹ x dx: one question using all the methods

This exam type comes up again and again. It uses almost every method in the chapter.

Take $u=\tan^{-1}x$ (you cannot integrate it) and $\dfrac{dv}{dx}=x$ , so $v=\dfrac{x^2}{2}$ . By parts:

$\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x - \int \frac{x^2}{2}\cdot\frac{1}{x^2+1}\,dx$

The leftover uses the $\tfrac{1}{x^2+1}$ derivative from §8.2.
After a quick $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ tidy-up, it finishes as $x$ minus an arctan.

Worked example

9709/32 Nov 2025 Q4: $\int_0^1 x\tan^{-1}x\,dx$ : the chapter in one question

$\tan^{-1}x$ is present → it is $u$ ; $\dfrac{dv}{dx}=x$ , so $v=\dfrac{x^2}{2}$ and $\dfrac{du}{dx}=\dfrac{1}{x^2+1}$ .
By parts: $\left[\tfrac{x^2}{2}\tan^{-1}x\right]_0^1 - \tfrac12\int_0^1 \frac{x^2}{x^2+1}\,dx$ .
Tidy the leftover with $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ , so it integrates to $x-\tan^{-1}x$ .
Put it together: $\tfrac12\cdot\tfrac{\pi}{4} - \tfrac12\left[x-\tan^{-1}x\right]_0^1 = \tfrac{\pi}{8} - \tfrac12\left(1-\tfrac{\pi}{4}\right)$ .

Answer

\int_0^1 x\tan^{-1}x\,dx = \tfrac{\pi}{4} - \tfrac12

(9709/32 Nov 2025 Q4). One question, three chapter skills: the by-parts choice of $u$ , the arctan derivative (§8.2), and the $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ divide-first tidy-up.

Your turn— tap to reveal the worked answer (9709/32 Jun 2025 Q10(b))

The same move on the $x\tan^{-1}2x$ version: “Find the exact value of $\displaystyle\int_0^{0.5} x\tan^{-1}2x\,dx$ ” (also set in 9709/31 Jun 2024 Q10(b)). Take $u=\tan^{-1}2x$ so $\tfrac{du}{dx}=\tfrac{2}{1+4x^2}$ , then the same divide-first tidy-up on $\dfrac{x^2}{1+4x^2}$ finishes it. (9709/32 Jun 2025 Q10)

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Integration by parts: integrating a product

By parts is the product rule run backwards.
A product of two unrelated functions ( $x\,e^{x}$ , $x\ln x$ , $x\tan^{-1}x$ ) could only have come from differentiating a product, so this is the tool to use.
The most important choice is which factor you call $u$ .

The formula and the choice of u

$\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$ (textbook 8.6, Key Point 8.4)

You choose $u$ (the part you differentiate) and $\dfrac{dv}{dx}$ (the part you integrate).
Pick $u$ to be the part that gets simpler when you differentiate it, and that you could not easily integrate anyway.

A rule you can use without thinking: if $\ln x$ or $\tan^{-1}x$ is there, that is $u$ (you cannot integrate them nicely). Otherwise $u$ = the power of $x$ .

Common mistake

The wrong choice of

u

wastes the whole question. You cannot integrate

\ln x

to get

v

, so for

x\ln x

you must take

u=\ln x

. If your leftover integral comes out harder than the one you started with, that is the sign to swap.

Worked example

9709/31 Nov 2024 Q2: $\int_1^3 x^2\ln 3x\,dx$

$\ln 3x$ is present → take $u=\ln 3x$ , $\dfrac{dv}{dx}=x^2$ .
Then $\dfrac{du}{dx}=\dfrac{1}{x}$ and $v=\dfrac{x^3}{3}$ .
Apply: $\left[\dfrac{x^3}{3}\ln 3x\right]_1^3 - \int_1^3 \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx$ ; the leftover $\int \dfrac{x^2}{3}\,dx$ is now trivial.

Worked example

9709/33 Jun 2020 Q2: $\int_0^1 (2-x)e^{-2x}\,dx$ : the other u

Now there is no $\ln$ or $\tan^{-1}$ , so the rule says $u$ = the polynomial, and the exponential is the part you integrate.

Take $u=2-x$ , $\dfrac{dv}{dx}=e^{-2x}$ . Then $\dfrac{du}{dx}=-1$ and $v=-\tfrac12 e^{-2x}$ .
Apply the formula: $\left[-\tfrac12(2-x)e^{-2x}\right]_0^1 - \int_0^1 \left(-\tfrac12 e^{-2x}\right)(-1)\,dx$ .
The leftover $-\tfrac12\int_0^1 e^{-2x}\,dx=\left[\tfrac14 e^{-2x}\right]_0^1$ is now a one-line integral. Combine the two brackets.

Answer

\int_0^1 (2-x)e^{-2x}\,dx = \tfrac34 - \tfrac14 e^{-2}

(9709/33 Jun 2020 Q2). The choice is the whole question: with no log here, the $x$ -polynomial is the part that gets simpler when you differentiate, so it is $u$ .

Why “differentiate the part that gets simpler” is the right rule

∫ ln x dx: write it as 1 · ln x

Integrating $\ln x$ on its own looks impossible, until you see it is really a product.

Write $\ln x = 1\cdot\ln x$ , then take $u=\ln x$ and $\dfrac{dv}{dx}=1$ (so $v=x$ ):

$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + c$

(textbook Worked Example 8.17). The same $1\cdot(\dots)$ move handles $\int \tan^{-1}x\,dx$ .

Your turn— tap to reveal the worked answer (9709/31 Nov 2025 Q1)

Answer

\int_1^2 \ln 3x\,dx = \ln 12 - 1

(9709/31 Nov 2025 Q1). Same $1\cdot(\dots)$ move as $\int\ln x\,dx$ . The only extra step is folding $2\ln 6-\ln 3$ with log laws.

Applying it twice

When $u$ is a higher power of $x$ (like $x^2\sin x$ or $x^2 e^{x}$ ), one pass turns $x^2$ into $2x$ but still leaves a product behind.
Do by parts a second time on that leftover. The power drops to a constant and it finishes.

Common mistake

On the second pass, the whole job is the signs. The formula's

-\int v\,\tfrac{du}{dx}

term flips the sign each time. Keep track of the minus. Most lost marks here are one dropped sign, not a wrong method. (textbook Worked Example 8.18)

∫ x tan⁻¹ x dx: one question using all the methods

This exam type comes up again and again. It uses almost every method in the chapter.

Take $u=\tan^{-1}x$ (you cannot integrate it) and $\dfrac{dv}{dx}=x$ , so $v=\dfrac{x^2}{2}$ . By parts:

$\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x - \int \frac{x^2}{2}\cdot\frac{1}{x^2+1}\,dx$

The leftover uses the $\tfrac{1}{x^2+1}$ derivative from §8.2.
After a quick $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ tidy-up, it finishes as $x$ minus an arctan.

Worked example

9709/32 Nov 2025 Q4: $\int_0^1 x\tan^{-1}x\,dx$ : the chapter in one question

$\tan^{-1}x$ is present → it is $u$ ; $\dfrac{dv}{dx}=x$ , so $v=\dfrac{x^2}{2}$ and $\dfrac{du}{dx}=\dfrac{1}{x^2+1}$ .
By parts: $\left[\tfrac{x^2}{2}\tan^{-1}x\right]_0^1 - \tfrac12\int_0^1 \frac{x^2}{x^2+1}\,dx$ .
Tidy the leftover with $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ , so it integrates to $x-\tan^{-1}x$ .
Put it together: $\tfrac12\cdot\tfrac{\pi}{4} - \tfrac12\left[x-\tan^{-1}x\right]_0^1 = \tfrac{\pi}{8} - \tfrac12\left(1-\tfrac{\pi}{4}\right)$ .

Answer

\int_0^1 x\tan^{-1}x\,dx = \tfrac{\pi}{4} - \tfrac12

(9709/32 Nov 2025 Q4). One question, three chapter skills: the by-parts choice of $u$ , the arctan derivative (§8.2), and the $\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}$ divide-first tidy-up.

Your turn— tap to reveal the worked answer (9709/32 Jun 2025 Q10(b))