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Module 8 · L14 of 20 ~30 min ⚡ +90 XP available

Integrating $\dfrac{a}{a^2+x^2}$

Last lesson the denominator contained a difference. Today it contains a sum — and that single change switches the answer from arcsine to arctangent. The standard integral $\displaystyle\int \dfrac{a}{a^2+x^2}\,dx = \tan^{-1}\!\!\left(\tfrac{x}{a}\right)+C$ is just as important as its arcsine cousin. In this lesson you'll prove it, read the form instantly, and handle definite integrals confidently.

Today's hook — Before using any formula, recall $\dfrac{d}{dx}\tan^{-1}\!\!\left(\dfrac{x}{a}\right)$. If you can write it, the integral follows immediately. Jot it now — you'll check it after card 05.

0/5QUESTS

Recall — your gut answer first

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From Lesson 11, recall: what is $\dfrac{d}{dx}\tan^{-1}\!\!\left(\dfrac{x}{a}\right)$? Write it below without looking it up. Compare it with the arcsine derivative from last lesson.

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The key idea: from derivative to integral

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We know from Lesson 11 that:

$$\frac{d}{dx}\!\left[\tan^{-1}\!\!\left(\frac{x}{a}\right)\right] = \frac{a}{a^2 + x^2}$$

Reversing this derivative gives the integral:

$$\int \frac{a}{a^2 + x^2}\,dx = \tan^{-1}\!\!\left(\frac{x}{a}\right) + C$$

$\displaystyle\int \frac{a}{a^2+x^2}\,dx = \tan^{-1}\!\!\left(\frac{x}{a}\right)+C$

Note the $a$ in the numerator

The standard form has $a$ on top, not 1. This means if your numerator is 1, divide: $\dfrac{1}{a^2+x^2} = \dfrac{1}{a}\cdot\dfrac{a}{a^2+x^2}$.

No domain restriction needed

Unlike arcsine, the arctangent integral is defined for all real $x$ since $a^2 + x^2 > 0$ always holds when $a \ne 0$.

Special case $a = 1$

When $a = 1$: $\displaystyle\int \dfrac{1}{1+x^2}\,dx = \tan^{-1}(x)+C$. Memorise this base case.

What you'll master

Know

Key facts

$\displaystyle\int \dfrac{a}{a^2+x^2}\,dx = \tan^{-1}\!\!\left(\dfrac{x}{a}\right)+C$ is a standard integral
$\displaystyle\int \dfrac{1}{1+x^2}\,dx = \tan^{-1}(x)+C$ is the special case $a=1$
This result follows directly from differentiating $\tan^{-1}(x/a)$

Understand

Concepts

Why a sum $a^2 + x^2$ in the denominator (no square root) signals arctangent
How to handle numerators that are not exactly $a$
The difference between this form and the arcsine form from Lesson 13

Can do

Skills

Evaluate indefinite integrals of the form $\dfrac{k}{a^2+x^2}$
Rewrite the integrand to extract $a$ and match the standard form
Compute definite integrals using the arctangent primitive

Key terms

$\tan^{-1}(x/a)$The inverse tangent (arctangent) of $x/a$. Output is an angle in $(-\pi/2, \pi/2)$. Also written $\arctan(x/a)$.

Standard integralA pre-proved integral listed in the HSC reference sheet. Recognise the form, identify $a$, apply directly.

$a$ (scale parameter)A nonzero constant appearing squared in the denominator. Identifies as $a = \sqrt{\text{constant term}}$.

Sum in denominator$a^2 + x^2$ (no square root) signals arctangent. Contrast with $\sqrt{a^2-x^2}$ which signals arcsine.

Arctangent range$\tan^{-1}(x) \in (-\pi/2, \pi/2)$. In particular, $\tan^{-1}(1) = \pi/4$, $\tan^{-1}(0) = 0$, $\tan^{-1}(-1) = -\pi/4$.

Numerator adjustmentIf the numerator is $k \ne a$, write $\dfrac{k}{a^2+x^2} = \dfrac{k}{a}\cdot\dfrac{a}{a^2+x^2}$ before integrating.

The standard integral — proof and form

core concept

Proof from first principles. Apply the chain rule to $F(x) = \tan^{-1}\!\!\left(\dfrac{x}{a}\right)$:

$$F'(x) = \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a}$$

Simplify the compound fraction:

$$F'(x) = \frac{1}{a\left(1 + x^2/a^2\right)} = \frac{1}{a \cdot \frac{a^2 + x^2}{a^2}} = \frac{a}{a^2 + x^2}$$

Since $F'(x) = \dfrac{a}{a^2+x^2}$, reversing gives the standard integral:

$$\boxed{\int \frac{a}{a^2 + x^2}\,dx = \tan^{-1}\!\!\left(\frac{x}{a}\right) + C, \quad a \neq 0}$$

Handling numerator 1 instead of $a$: Often the integrand is $\dfrac{1}{a^2+x^2}$. Divide both sides by $a$:

$$\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\!\!\left(\frac{x}{a}\right) + C$$

HSC reference sheet. Both forms appear in the standard integrals table. Know how to use each: if the numerator matches $a$ exactly, the answer is $\tan^{-1}(x/a)+C$. If the numerator is any other constant $k$, write $k/a$ in front.

Proof from first principles. Apply the chain rule to $F(x) = \tan^{-1}\!\!\left(\dfrac{x}{a}\right)$:

Pause — copy the standard integral $\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}\tan^{-1}(x/a)+C$ with the full chain-rule proof into your book.

Quick check: What is $\displaystyle\int \dfrac{3}{9 + x^2}\,dx$?

Spotting the form — reading the integrand

core concept

We just saw the proof: $\frac{d}{dx}\tan^{-1}(x/a)=\frac{a}{a^2+x^2}$, so $\int\frac{a}{a^2+x^2}\,dx=\tan^{-1}(x/a)+C$, or equivalently $\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}\tan^{-1}(x/a)+C$. That raises a question: how do you identify the value of $a$ in a given integrand, and what does it look like when the coefficient of $x^2$ in the denominator is not 1? This card answers it → write denominator as $a^2+x^2$ by factoring; if coefficient $\neq1$, factor it out first.

The standard form is $\dfrac{a}{a^2 + x^2}$. To apply it:

Identify $a^2$: The constant term in the denominator is $a^2$. Extract $a = \sqrt{\text{constant}}$.
Check the numerator:
- If numerator $= a$: answer is $\tan^{-1}(x/a)+C$.
- If numerator $= 1$: answer is $\dfrac{1}{a}\tan^{-1}(x/a)+C$.
- If numerator $= k$: write $\dfrac{k}{a} \cdot \dfrac{a}{a^2+x^2}$ and integrate to get $\dfrac{k}{a}\tan^{-1}(x/a)+C$.
Check coefficient of $x^2$ is 1: If not, factor it out first to expose $a^2+x^2$.

Example — numerator matches $a$: $\displaystyle\int \dfrac{5}{25+x^2}\,dx$. Here $a^2 = 25$, $a = 5$, numerator $= 5 = a$. Answer: $\tan^{-1}\!\!\left(\dfrac{x}{5}\right)+C$.

Example — numerator is 1: $\displaystyle\int \dfrac{1}{4+x^2}\,dx$. Here $a = 2$, numerator $= 1$. Answer: $\dfrac{1}{2}\tan^{-1}\!\!\left(\dfrac{x}{2}\right)+C$.

Contrast with Lesson 13. Arcsine: $\sqrt{a^2 - x^2}$ (square root, difference). Arctangent: $a^2 + x^2$ (no root, sum). Distinguishing these two at a glance is a key exam skill.

The standard form is $\dfrac{a}{a^2 + x^2}$. To apply it:

Pause — copy the form-recognition steps: (1) write denominator as $a^2+x^2$; (2) identify $a$; (3) apply $\frac{1}{a}\tan^{-1}(x/a)+C$ into your book.

Did you get this? True or false: $\displaystyle\int \dfrac{1}{4+x^2}\,dx = \dfrac{1}{2}\tan^{-1}\!\!\left(\dfrac{x}{2}\right)+C$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · NUMERATOR = a

Find $\displaystyle\int \dfrac{2}{4+x^2}\,dx$.

Identify $a$: denominator is $4 + x^2 = 2^2 + x^2$, so $a = 2$.

Extract $a$ from the constant term. Here $a^2 = 4$ gives $a = 2$.

PROBLEM 2 · NUMERATOR = 1

Find $\displaystyle\int \dfrac{1}{9+x^2}\,dx$.

$a^2 = 9$ so $a = 3$. Numerator is 1, not 3. Use the divided form.

Since numerator $\ne a$, write $\dfrac{1}{9+x^2} = \dfrac{1}{3}\cdot\dfrac{3}{9+x^2}$ to match the standard form.

PROBLEM 3 · DEFINITE INTEGRAL

Evaluate $\displaystyle\int_{0}^{5} \dfrac{5}{25+x^2}\,dx$. Give an exact answer.

$a^2 = 25$, $a = 5$. Numerator $= 5 = a$. Primitive: $F(x) = \tan^{-1}\!\!\left(\dfrac{x}{5}\right)$.

Standard form applies directly. Find the primitive first, then evaluate between limits.

Fill the gap: $\displaystyle\int \dfrac{1}{25+x^2}\,dx = \dfrac{1}{\,\,}\tan^{-1}\!\!\left(\dfrac{x}{5}\right)+C$. The coefficient should be .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to divide by $a$ when numerator is 1

$\displaystyle\int \dfrac{1}{9+x^2}\,dx = \dfrac{1}{3}\tan^{-1}(x/3)+C$, NOT $\tan^{-1}(x/3)+C$. When the numerator does not equal $a$, a coefficient of $1/a$ is needed. Always check: differentiate your answer — you must recover the integrand.

Trap 02

Writing $\tan^{-1}(ax)$ instead of $\tan^{-1}(x/a)$

The argument is $x/a$, not $ax$. For $\displaystyle\int \dfrac{3}{9+x^2}\,dx$: $a = 3$, answer is $\tan^{-1}(x/3)+C$. Writing $\tan^{-1}(3x)$ gives the wrong derivative — it produces $\dfrac{3}{1+9x^2}$, not $\dfrac{3}{9+x^2}$.

Trap 03

Confusing this with the arcsine form

Memory device: Sum, no root → arctan; Difference with root → arcsin. $\dfrac{1}{a^2+x^2}$ has no square root and uses $+$: that is arctangent. $\dfrac{1}{\sqrt{a^2-x^2}}$ has a square root and uses $-$: that is arcsine.

Did you get this? True or false: $\displaystyle\int \dfrac{1}{1+x^2}\,dx = \tan^{-1}(x)+C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find $\displaystyle\int \dfrac{7}{49+x^2}\,dx$.

Find $\displaystyle\int \dfrac{1}{1+x^2}\,dx$.

Evaluate $\displaystyle\int_{0}^{3} \dfrac{3}{9+x^2}\,dx$. Give an exact answer.

Verify that $\dfrac{d}{dx}\tan^{-1}\!\!\left(\dfrac{x}{4}\right) = \dfrac{4}{16+x^2}$ using the chain rule.

Find $\displaystyle\int \dfrac{6}{a^2+x^2}\,dx$ in terms of $a$.

Odd one out: Three of these integrals give an arctangent result. Which one does NOT?

Revisit your thinking

Earlier you wrote down what you thought $\dfrac{d}{dx}\tan^{-1}(x/a)$ equals.

The exact answer is $\dfrac{a}{a^2+x^2}$. This is precisely the integrand we integrated all lesson. Now you have both standard inverse trig integrals: arcsine from Lesson 13 and arctangent from this lesson. The key distinction is: difference with square root → arcsine; sum without square root → arctangent. Keeping this contrast sharp will save you time in any exam.

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Multiple choice

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Short answer

ApplyBand 31 mark

Q1. Find $\displaystyle\int \dfrac{1}{1+x^2}\,dx$. (1 mark)

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ApplyBand 42 marks

Q2. Evaluate $\displaystyle\int_{0}^{5} \dfrac{5}{25+x^2}\,dx$. Give an exact answer. (2 marks)

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AnalyseBand 52 marks

Q3. Show that $\dfrac{d}{dx}\tan^{-1}\!\!\left(\dfrac{x}{a}\right) = \dfrac{a}{a^2+x^2}$ using the chain rule. Hence write down $\displaystyle\int \dfrac{6}{a^2+x^2}\,dx$. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\tan^{-1}(x/7)+C$ · 2. $\tan^{-1}(x)+C$ · 3. $[\tan^{-1}(x/3)]_0^3 = \tan^{-1}(1) - 0 = \pi/4$ · 4. $u = x/4$; $\dfrac{d}{dx}\tan^{-1}(x/4) = \dfrac{1}{1+(x/4)^2}\cdot\dfrac{1}{4} = \dfrac{4}{16+x^2}$ ✓ · 5. $\dfrac{6}{a}\tan^{-1}(x/a)+C$

Q1 (1 mark): $\displaystyle\int \dfrac{1}{1+x^2}\,dx = \tan^{-1}(x)+C$ [1].

Q2 (2 marks): $a = 5$; numerator $= 5 = a$; primitive $= \tan^{-1}(x/5)$ [1]. $[\tan^{-1}(x/5)]_0^5 = \tan^{-1}(1) - \tan^{-1}(0) = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}$ [1].

Q3 (2 marks): Let $u = x/a$; $\dfrac{d}{dx}\tan^{-1}(u) = \dfrac{1}{1+u^2}\cdot\dfrac{1}{a} = \dfrac{1}{a(1+x^2/a^2)} = \dfrac{a}{a^2+x^2}$ [1]. Hence $\displaystyle\int \dfrac{6}{a^2+x^2}\,dx = \dfrac{6}{a}\tan^{-1}\!\!\left(\dfrac{x}{a}\right)+C$ [1].

Boss battle · The Integral Master

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← Lesson 13 · Integrating 1/√(a²−x²) Lesson 15 · Completing the Square →

Module overview · Extension 1 · Checkpoint 3