Mathematics 9709 · Pure Mathematics 3

The arctan integral: 1/(x²+a²) and splitting (linear)/(x²+a²)

Some integrands have nothing to factor and no obvious $\tfrac{f'}{f}$ shape: a smooth bump like $\dfrac{1}{x^2+a^2}$ .
Run the backwards test (what differentiates to that?). The answer is $\tan^{-1}$ , so this whole family integrates back to an arctan.

Why arctan: the derivative that produces it

Take $y=\tan^{-1}x$ , i.e. $\tan y = x$ . Differentiate implicitly: $\sec^2 y\,\dfrac{dy}{dx}=1$ , and since $\sec^2 y = 1+\tan^2 y = 1+x^2$ ,

$\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{x^2+1}$ (textbook 8.1, Key Point 8.1)

You rarely get asked to just differentiate $\tan^{-1}(\dots)$ on its own. Its real job is to set up the integral below.

y = 1/(x² + 4): a smooth bump with nothing to factor. It came from an arctan, so it integrates back to one arctan.

The standard integral

$\displaystyle\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + c$ (textbook 8.2, Key Point 8.2)

Out front it is $\tfrac{1}{a}$ (not $\tfrac{1}{a^2}$ ).
The inside is $\tfrac{x}{a}$ (not $\tfrac{x}{a^2}$ ).
So $a$ is the square root of the constant under the bump.

Worked example

Find $\int \frac{1}{x^2+25}\,dx$

Here $a^2=25$ so $a=5$ .

Answer

\tfrac{1}{5}\tan^{-1}\!\frac{x}{5} + c

(textbook Worked Example 8.2)

Common mistake

If the

x^2

has a number in front, factor it out first so the bottom reads

x^2+a^2

. For

\dfrac{1}{2x^2+3}

, write

\dfrac{1}{2}\cdot\dfrac{1}{x^2+\frac{3}{2}}

. Now

a^2=\tfrac32

Where the 1/a out front comes from

The $\tfrac1a$ is not a rule to memorise. It drops out of the chain rule. The answer $\tan^{-1}\tfrac{x}{a}$ has an inside function $\tfrac{x}{a}$ , whose derivative is $\tfrac1a$ . So differentiating $\tan^{-1}\tfrac{x}{a}$ gives $\dfrac{1}{1+(x/a)^2}\cdot\dfrac1a=\dfrac{a}{x^2+a^2}$ : an extra factor of $a$ . To cancel it you divide by $a$ , and that divide is the $\tfrac1a$ sitting out front. It also explains why the inside is $\tfrac{x}{a}$ and not $\tfrac{x}{a^2}$ : $a$ is the square root of the constant, not the constant itself.

The key skill: splitting (linear)/(x² + a²)

When you see $\dfrac{\text{linear}+\text{const}}{x^2+a^2}$ , it is two different integrals stuck together, so split the top:

$\frac{px+q}{x^2+a^2} = \underbrace{\frac{px}{x^2+a^2}}_{\to\ \frac{p}{2}\ln(x^2+a^2)} + \underbrace{\frac{q}{x^2+a^2}}_{\to\ \frac{q}{a}\tan^{-1}(x/a)}$

The $x$ -part is a $\tfrac{f'}{f}$ form (§8.1) → $\ln$ . The constant part is the standard form → arctan. Do each one on its own, then add them.

Worked example

9709/32 Nov 2023 Q5: $\int_0^6 \frac{x(x+1)}{x^2+4}\,dx$

Top is $x(x+1)=x^2+x$ — same degree as the bottom, so the fraction is improper. Divide first: $\dfrac{x^2+x}{x^2+4}=1+\dfrac{x-4}{x^2+4}$ .
Now split the proper remainder: $\dfrac{x-4}{x^2+4}=\dfrac{x}{x^2+4}-\dfrac{4}{x^2+4}$ .
Integrate the three pieces. The $1$ gives $x$ ; the $x$ -piece is $\tfrac{f'}{f}$ → $\tfrac12\ln(x^2+4)$ ; the constant piece is the standard form with $a=2$ → $-4\cdot\tfrac12\tan^{-1}\!\frac{x}{2}=-2\tan^{-1}\!\frac{x}{2}$ .
Antiderivative: $\left[\,x+\tfrac12\ln(x^2+4)-2\tan^{-1}\!\tfrac{x}{2}\,\right]_0^6$ . At $x=6$ : $6+\tfrac12\ln 40-2\tan^{-1}3$ ; at $x=0$ : $\tfrac12\ln 4$ . Subtract, and $\tfrac12\ln 40-\tfrac12\ln 4=\tfrac12\ln 10$ .

Answer

\int_0^6 \frac{x(x+1)}{x^2+4}\,dx = 6 + \tfrac12\ln 10 - 2\tan^{-1}3

(9709/32 Nov 2023 Q5). The question asks for an exact value, so leave it as that $\ln$ -and- $\tan^{-1}$ expression.

Common mistake

The big error is integrating only one of the two pieces, usually the arctan and forgetting the

\ln

, or the other way round. The other is leaving the answer as a decimal when it says “exact”. The answer should be an exact multiple of

\pi

, or an

a+b\ln c

form.

The given trig substitution variant

CAIE sometimes gives you $x=a\tan\theta$ for a harder arctan-family integral. You do not invent it; they give it.
How the substitution works is in §8.3. Here, just spot the $x=a\tan\theta$ cue as another arctan-family integral.

Your turn— tap to reveal the worked answer (9709/31 Jun 2022 Q6)

“Using the substitution $x=3\tan\theta$ , show that $I=\displaystyle\int_0^3 \frac{27}{(9+x^2)^2}\,dx = \int_0^{\pi/4} \cos^2\theta\,d\theta$ , hence find the exact value.” The $x=3\tan\theta$ turns $9+x^2$ into $9\sec^2\theta$ . With $dx=3\sec^2\theta\,d\theta$ , it tidies down to a $\cos^2\theta$ integral, with no leftover factor out front. The exact value is $\tfrac14+\tfrac{\pi}{8}$ . (9709/31 Jun 2022 Q6)

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The arctan integral: 1/(x²+a²) and splitting (linear)/(x²+a²)

Some integrands have nothing to factor and no obvious $\tfrac{f'}{f}$ shape: a smooth bump like $\dfrac{1}{x^2+a^2}$ .
Run the backwards test (what differentiates to that?). The answer is $\tan^{-1}$ , so this whole family integrates back to an arctan.

Why arctan: the derivative that produces it

Take $y=\tan^{-1}x$ , i.e. $\tan y = x$ . Differentiate implicitly: $\sec^2 y\,\dfrac{dy}{dx}=1$ , and since $\sec^2 y = 1+\tan^2 y = 1+x^2$ ,

$\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{x^2+1}$ (textbook 8.1, Key Point 8.1)

You rarely get asked to just differentiate $\tan^{-1}(\dots)$ on its own. Its real job is to set up the integral below.

y = 1/(x² + 4): a smooth bump with nothing to factor. It came from an arctan, so it integrates back to one arctan.

The standard integral

$\displaystyle\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + c$ (textbook 8.2, Key Point 8.2)

Out front it is $\tfrac{1}{a}$ (not $\tfrac{1}{a^2}$ ).
The inside is $\tfrac{x}{a}$ (not $\tfrac{x}{a^2}$ ).
So $a$ is the square root of the constant under the bump.

Worked example

Find $\int \frac{1}{x^2+25}\,dx$

Here $a^2=25$ so $a=5$ .

Answer

\tfrac{1}{5}\tan^{-1}\!\frac{x}{5} + c

(textbook Worked Example 8.2)

Common mistake

If the

x^2

has a number in front, factor it out first so the bottom reads

x^2+a^2

. For

\dfrac{1}{2x^2+3}

, write

\dfrac{1}{2}\cdot\dfrac{1}{x^2+\frac{3}{2}}

. Now

a^2=\tfrac32

Where the 1/a out front comes from

The key skill: splitting (linear)/(x² + a²)

When you see $\dfrac{\text{linear}+\text{const}}{x^2+a^2}$ , it is two different integrals stuck together, so split the top:

$\frac{px+q}{x^2+a^2} = \underbrace{\frac{px}{x^2+a^2}}_{\to\ \frac{p}{2}\ln(x^2+a^2)} + \underbrace{\frac{q}{x^2+a^2}}_{\to\ \frac{q}{a}\tan^{-1}(x/a)}$

The $x$ -part is a $\tfrac{f'}{f}$ form (§8.1) → $\ln$ . The constant part is the standard form → arctan. Do each one on its own, then add them.

Worked example

9709/32 Nov 2023 Q5: $\int_0^6 \frac{x(x+1)}{x^2+4}\,dx$

Top is $x(x+1)=x^2+x$ — same degree as the bottom, so the fraction is improper. Divide first: $\dfrac{x^2+x}{x^2+4}=1+\dfrac{x-4}{x^2+4}$ .
Now split the proper remainder: $\dfrac{x-4}{x^2+4}=\dfrac{x}{x^2+4}-\dfrac{4}{x^2+4}$ .
Integrate the three pieces. The $1$ gives $x$ ; the $x$ -piece is $\tfrac{f'}{f}$ → $\tfrac12\ln(x^2+4)$ ; the constant piece is the standard form with $a=2$ → $-4\cdot\tfrac12\tan^{-1}\!\frac{x}{2}=-2\tan^{-1}\!\frac{x}{2}$ .
Antiderivative: $\left[\,x+\tfrac12\ln(x^2+4)-2\tan^{-1}\!\tfrac{x}{2}\,\right]_0^6$ . At $x=6$ : $6+\tfrac12\ln 40-2\tan^{-1}3$ ; at $x=0$ : $\tfrac12\ln 4$ . Subtract, and $\tfrac12\ln 40-\tfrac12\ln 4=\tfrac12\ln 10$ .

Answer

\int_0^6 \frac{x(x+1)}{x^2+4}\,dx = 6 + \tfrac12\ln 10 - 2\tan^{-1}3

(9709/32 Nov 2023 Q5). The question asks for an exact value, so leave it as that $\ln$ -and- $\tan^{-1}$ expression.

Common mistake

The big error is integrating only one of the two pieces, usually the arctan and forgetting the

\ln

, or the other way round. The other is leaving the answer as a decimal when it says “exact”. The answer should be an exact multiple of

\pi

, or an

a+b\ln c

form.

The given trig substitution variant

CAIE sometimes gives you $x=a\tan\theta$ for a harder arctan-family integral. You do not invent it; they give it.
How the substitution works is in §8.3. Here, just spot the $x=a\tan\theta$ cue as another arctan-family integral.

Your turn— tap to reveal the worked answer (9709/31 Jun 2022 Q6)