Integration by parts | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Rearrange the product rule

Let $u$ and $v$ be functions of $x$ . The product rule says:

\frac d{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

Integrating both sides and moving the second integral gives the by-parts formula. In the formula, $dv/dx$ is the factor you integrate to obtain $v$ .

\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx

Choose roles that make the next integral easier

Give each factor one job

differentiate

choose what becomes simpler

dv/dx

integrate to get v

choose what integrates directly

∫u(dv/dx) dx = uv

− ∫v(du/dx) dx

If the new integral is harder, swap the roles.

Choose roles by what becomes easier, not by which factor appears first.

Choose $u$ as a logarithm or inverse tangent if one is present; otherwise a polynomial is often the factor that shrinks.
Choose $dv/dx$ as the factor you can integrate directly to get $v$ .
If the new integral is harder than the original, reconsider the roles.

Common mistake

The formula has a minus sign. Write the complete

uv-\int v\,du/dx\,dx

line before simplifying.

Keep parameters symbolic in a current-paper integral

Worked example

9709/32/M/J/24 Q6(b)

\int_0^{2/a}xe^{-ax}\,dx,\qquad a>0

Choose $u=x$ and $dv/dx=e^{-ax}$ . Then $du/dx=1$ and $v=-e^{-ax}/a$ .

\int xe^{-ax}\,dx

=-\frac xa e^{-ax}-\frac1{a^2}e^{-ax}

Apply the limits only after the antiderivative is complete.

Answer

\frac1{a^2}\left(1-3e^{-2}\right)

(9709/32/M/J/24 Q6(b))

Examiner note

A parameter such as

a

behaves like a constant. Do not replace it with a number or differentiate it.

Recognise one-factor products and repeated use

For $\int\ln x\,dx$ , write $1\cdot\ln x$ and choose $u=\ln x$ . This gives $x\ln x-x+C$ .
The same hidden product works for $\int\tan^{-1}x\,dx$ . Choose $u=\tan^{-1}x$ and $dv=dx$ to get:
Answer
$x\tan^{-1}x$
$-\frac12\ln(1+x^2)+C$
For products such as $x^2e^x$ , the first pass leaves another product. Apply integration by parts again.

Key idea

By parts is successful when each pass reduces the difficult factor: for example,

x^2\to2x\to2

Worked example

Use by parts twice

\int x^2e^x\,dx

First choose $u=x^2$ and $dv/dx=e^x$ :

=x^2e^x-2\int xe^x\,dx

The remaining product still has a polynomial factor, so use by parts again:

\int xe^x\,dx=xe^x-e^x

Answer

e^x(x^2-2x+2)+C

(Pure Mathematics 2 & 3 Coursebook, Ch. 8.6–8.7)

Your turnIndependent check [5 marks]

Find the exact value of $\displaystyle\int_0^1xe^{-2x}\,dx$ .

Show worked answer

Choose $u=x$ and $dv/dx=e^{-2x}$ , so $v=-\frac12e^{-2x}$ .

\int xe^{-2x}\,dx

=-\frac x2e^{-2x}-\frac14e^{-2x}

Answer

\frac14-\frac{3}{4e^2}

Rearrange the product rule

Let $u$ and $v$ be functions of $x$ . The product rule says:

\frac d{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

Integrating both sides and moving the second integral gives the by-parts formula. In the formula, $dv/dx$ is the factor you integrate to obtain $v$ .

\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx

Choose roles that make the next integral easier

Give each factor one job

differentiate

choose what becomes simpler

dv/dx

integrate to get v

choose what integrates directly

∫u(dv/dx) dx = uv

− ∫v(du/dx) dx

If the new integral is harder, swap the roles.

Choose roles by what becomes easier, not by which factor appears first.

Choose $u$ as a logarithm or inverse tangent if one is present; otherwise a polynomial is often the factor that shrinks.
Choose $dv/dx$ as the factor you can integrate directly to get $v$ .
If the new integral is harder than the original, reconsider the roles.

Common mistake

The formula has a minus sign. Write the complete

uv-\int v\,du/dx\,dx

line before simplifying.

Keep parameters symbolic in a current-paper integral

Worked example

9709/32/M/J/24 Q6(b)

\int_0^{2/a}xe^{-ax}\,dx,\qquad a>0

Choose $u=x$ and $dv/dx=e^{-ax}$ . Then $du/dx=1$ and $v=-e^{-ax}/a$ .

\int xe^{-ax}\,dx

=-\frac xa e^{-ax}-\frac1{a^2}e^{-ax}

Apply the limits only after the antiderivative is complete.

Answer

\frac1{a^2}\left(1-3e^{-2}\right)

(9709/32/M/J/24 Q6(b))

Examiner note

A parameter such as

a

behaves like a constant. Do not replace it with a number or differentiate it.

Recognise one-factor products and repeated use

For $\int\ln x\,dx$ , write $1\cdot\ln x$ and choose $u=\ln x$ . This gives $x\ln x-x+C$ .
The same hidden product works for $\int\tan^{-1}x\,dx$ . Choose $u=\tan^{-1}x$ and $dv=dx$ to get:
Answer
$x\tan^{-1}x$
$-\frac12\ln(1+x^2)+C$
For products such as $x^2e^x$ , the first pass leaves another product. Apply integration by parts again.

Key idea

By parts is successful when each pass reduces the difficult factor: for example,

x^2\to2x\to2

Worked example

Use by parts twice

\int x^2e^x\,dx

First choose $u=x^2$ and $dv/dx=e^x$ :

=x^2e^x-2\int xe^x\,dx

The remaining product still has a polynomial factor, so use by parts again:

\int xe^x\,dx=xe^x-e^x

Answer

e^x(x^2-2x+2)+C

(Pure Mathematics 2 & 3 Coursebook, Ch. 8.6–8.7)

Your turnIndependent check [5 marks]

Find the exact value of $\displaystyle\int_0^1xe^{-2x}\,dx$ .

Show worked answer

Choose $u=x$ and $dv/dx=e^{-2x}$ , so $v=-\frac12e^{-2x}$ .

\int xe^{-2x}\,dx

=-\frac x2e^{-2x}-\frac14e^{-2x}

Answer

\frac14-\frac{3}{4e^2}