Chapter 7 · Integrals

Exercise 7.2 · Q1: $\int \frac{2x}{1+x^2}\,dx$

Exercise 7.2Integration by substitution · Section 7.3.1

The question

Integrate the function:

\frac{2x}{1+x^2}

Exam answer

Let $I = \displaystyle\int\frac{2x}{1 + x^2}\,dx$ ⁠.

Put $1 + x^2 = t$ ⁠. Then $2x\,dx = dt$ ⁠.

\therefore\ I = \int\frac{dt}{t} = \log|t| + C

$\therefore\ I = \log\left(1 + x^2\right) + C$ (as $1 + x^2 > 0$ ⁠).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Look at the denominator $1+x^2$ ⁠. Its derivative is $2x$ ⁠, which is exactly the numerator. Whenever the top of a fraction is the derivative of the bottom, the substitution "bottom $= t$ " turns the integral into $\int \frac{dt}{t}$ ⁠, which is $\log|t|$ ⁠. So the answer is the log of the denominator.

Given, to find, and the plan

Given: The integrand $f(x) = \dfrac{2x}{1+x^2}$ ⁠. Its denominator satisfies $1+x^2 \ge 1 > 0$ ⁠, so $f$ is defined and continuous for every real $x$ ⁠.
To find: $\displaystyle\int \frac{2x}{1+x^2}\,dx$ ⁠, that is, every function whose derivative is $f(x)$ ⁠.
Plan: (1) Notice that the numerator is the derivative of the denominator. (2) Substitute $t = 1+x^2$ ⁠. (3) Integrate $\frac1t$ with the standard formula. (4) Return to $x$ ⁠. (5) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

$\int f(x)\,dx = F(x) + C$ means exactly that $F'(x) = f(x)$ ⁠. So any answer can be checked by differentiating it. Basics: integration undoes differentiation

2Substitution

If we put $t = g(x)$ ⁠, then $dt = g'(x)\,dx$ ⁠, and $\int h(g(x))\,g'(x)\,dx = \int h(t)\,dt$ ⁠. It is the chain rule read backwards (Section 7.3.1). Basics: substitution

3The integral of $1/t$

$\int \frac{1}{t}\,dt = \log|t| + C$ for $t \ne 0$ (standard formula (xii) in Section 7.2). The absolute value is there because $\log$ only accepts positive numbers. Basics: logarithms and 1/x

Solution

Spot the derivative of the denominator in the numerator
$\frac{d}{dx}\left(1+x^2\right) = 0 + 2x = 2x.$
So the integrand has the shape $\dfrac{g'(x)}{g(x)}$ with $g(x) = 1+x^2$ ⁠.

Substitution works when the integrand contains some inner expression together with its derivative. Here the inner expression is the denominator and its derivative $2x$ is sitting in the numerator, so the substitution $t = 1+x^2$ will absorb the whole numerator into $dt$ ⁠.

$\frac{d}{dx}(x^2) = 2x$ (power rule) and a constant has derivative $0$ ⁠. Basics: powers of x
Substitute $t = 1+x^2$
$\begin{aligned} t &= 1+x^2, \\ \frac{dt}{dx} &= 2x, \\ dt &= 2x\,dx. \end{aligned}$
The integral becomes
$\int \frac{2x}{1+x^2}\,dx = \int \frac{1}{1+x^2}\,(2x\,dx) = \int \frac{dt}{t}.$
We split the integrand as $\frac{1}{1+x^2}$ times $2x\,dx$ ⁠. The first factor is $\frac1t$ and the second is exactly $dt$ ⁠. Nothing in $x$ is left over, which tells us the substitution was the right one. Note that $t = 1+x^2 \ge 1$ ⁠, so $t$ is never $0$ and $\frac1t$ is always defined.

Writing $dt = g'(x)\,dx$ is the shorthand of Section 7.3.1: "put $x = g(t)$ so that $dx = g'(t)\,dt$ ", used here in the direction $t = g(x)$ ⁠. Lesson: integration by substitution
Integrate in $t$
$\int \frac{dt}{t} = \log|t| + C.$
This is the standard formula $\int \frac1x\,dx = \log|x| + C$ ⁠, with the letter $t$ in place of $x$ (Remark (iii) in Section 7.2 says the formulas hold for any name of the variable).

Standard formula (xii), Section 7.2: $\frac{d}{dx}\log|x| = \frac1x$ ⁠, so $\int\frac1x\,dx = \log|x| + C$ ⁠. Basics: logarithms and 1/x
Return to $x$ and drop the absolute value
$\log|t| + C = \log\left|1+x^2\right| + C = \log\left(1+x^2\right) + C.$
The question is in $x$ ⁠, so the answer must be in $x$ ⁠: replace $t$ by $1+x^2$ ⁠. Because $1+x^2$ is always positive, $|1+x^2| = 1+x^2$ and the absolute value bars can be removed. The answer is valid for every real $x$ ⁠.
Check: differentiate the answer

Use the chain rule with inner function $u = 1+x^2$ and outer function $\log u$ ⁠:
$\begin{aligned} \frac{d}{dx}\left[\log(1+x^2) + C\right] &= \frac{1}{1+x^2}\cdot\frac{d}{dx}(1+x^2) + 0 \\ &= \frac{1}{1+x^2}\cdot 2x \\ &= \frac{2x}{1+x^2}. \end{aligned}$
This is the integrand, so the answer is correct.

An indefinite integral is right exactly when its derivative gives back the integrand (Property (I), Section 7.2.1). Differentiating is always easier than integrating, so this check costs little and catches slips such as a missing factor.

$\frac{d}{du}\log u = \frac1u$ (Chapter 5, Theorem 5) and the chain rule (Chapter 5, Theorem 4). Basics: integration undoes differentiation

Result

For every real $x$ ⁠,

\int \frac{2x}{1+x^2}\,dx = \log\left(1+x^2\right) + C.

(Textbook answer: $\log(1+x^2) + C$ ⁠.)

Check it with numbers

Let $F(x) = \log(1+x^2)$ ⁠. At $x = 2$ the integrand is $\frac{2\cdot 2}{1+4} = 0.8$ ⁠. Now estimate the slope of $F$ at $2$ with $h = 0.0001$ ⁠:

\begin{aligned} F(2.0001) &\approx 1.60951791, \\ F(1.9999) &\approx 1.60935791, \\ \frac{F(2.0001)-F(1.9999)}{0.0002} &\approx 0.800000. \end{aligned}

The slope of our answer equals the integrand, as it should.

Common mistakes

Writing $\log(1+x^2)$ times something extra. Once $2x\,dx$ has become $dt$ ⁠, the factor $2x$ is used up. It does not appear again in the answer.
Integrating the top and bottom separately. $\int \frac{2x}{1+x^2}\,dx$ is not $\frac{x^2}{x + x^3/3}$ ⁠. There is no "quotient rule" for integrals.
Thinking the absolute value is wrong. The general formula has $|t|$ ⁠. Here it can be dropped only because $1+x^2 > 0$ ⁠; say so.
Forgetting $+\,C$ ⁠. The integral is a whole family of functions.

What to take away

Whenever the numerator is the derivative of the denominator, $\int \frac{g'(x)}{g(x)}\,dx = \log|g(x)| + C$ ⁠.
Choose the substitution so that $dt$ swallows the rest of the integrand exactly.
Always differentiate your answer to check it.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}x^2 = 2x$ ⁠; a constant has derivative 0	Steps 1, 2	Chapter 5, Table 5.3
Substitution $t = g(x)$ ⁠, $dt = g'(x)\,dx$	Step 2	Section 7.3.1 (Lesson: substitution)
$\int \frac1t\,dt = \log\|t\| + C$	Step 3	Standard formula (xii), Section 7.2
$\|a\| = a$ when $a > 0$	Step 4	Definition of absolute value
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Step 5	Chapter 5, Theorem 5 and Theorem 4
An integral is checked by differentiating it	Step 5	Property (I), Section 7.2.1

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Exercise 7.2 · Q1: ∫2x1+x2 dx\int \frac{2x}{1+x^2}\,dx∫1+x22x​dx

Basics you need

Solution

Spot the derivative of the denominator in the numerator

Substitute t=1+x2t = 1+x^2t=1+x2

Integrate in ttt

Return to xxx and drop the absolute value

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.2 · Q1: $\int \frac{2x}{1+x^2}\,dx$

Substitute $t = 1+x^2$

Integrate in $t$

Return to $x$ and drop the absolute value