Chapter 7 · Integrals

Exercise 7.9 · Q1: $\int_0^1\frac{x}{x^2+1}\,dx$

Exercise 7.9Definite integrals by substitution · Section 7.9

The question

Evaluate the integral using substitution:

\int_0^1 \frac{x}{x^2 + 1}\,dx.

Exam answer

Let $I = \displaystyle\int_0^1 \frac{x}{x^2 + 1}\,dx$ ⁠.

Put $x^2 + 1 = t$ ⁠. Then $2x\,dx = dt$ ⁠, so $x\,dx = \frac12\,dt$ ⁠.

When $x = 0$ ⁠, $t = 1$ ⁠; when $x = 1$ ⁠, $t = 2$ ⁠.

\begin{aligned} I &= \frac12\int_1^2\frac{dt}{t} = \frac12\Big[\log t\Big]_1^2 \\ &= \frac12\left(\log 2 - \log 1\right) \end{aligned}

$\therefore\ I = \dfrac12\log 2$ ⁠.

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

The one idea this problem is built on

The numerator $x$ is half the derivative of the denominator $x^2 + 1$ ⁠. So with $t = x^2 + 1$ we get $x\,dx = \frac{1}{2}\,dt$ ⁠, and the integral becomes $\frac{1}{2}\int\frac{dt}{t}$ ⁠, a logarithm. The limits move with the substitution: $x = 0$ gives $t = 1$ and $x = 1$ gives $t = 2$ ⁠.

Given, to find, and the plan

Given: The integrand $\frac{x}{x^2 + 1}$ on the interval $[0, 1]$ ⁠.
To find: The exact value of the integral, and a decimal.
Plan: (1) Check the integrand is continuous on $[0, 1]$ ⁠. (2) Substitute $t = x^2 + 1$ ⁠, write $dt$ ⁠, and change the limits. (3) Integrate $\frac{1}{t}$ and evaluate. (4) Repeat with an antiderivative in $x$ ⁠, checked by differentiating. (5) Look at the region and compare with a numerical value.

Basics you need

1Substitution with new limits

If $t = g(x)$ has a continuous derivative on $[a, b]$ and $f$ is continuous on the values $g$ takes, then $\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt.$ This is the quick method in the Note of Section 7.9. Lesson: integration by substitution

2The integral of $\frac{1}{t}$

$\int\frac{dt}{t} = \log|t| + C$ for $t \ne 0$ (standard formula (xii), Section 7.2). Basics: logarithms and 1/x

3The second fundamental theorem

If $f$ is continuous on $[a, b]$ and $F' = f$ ⁠, then $\int_a^b f(x)\,dx = F(b) - F(a)$ (Theorem 2, Section 7.8.3). Lesson: definite integrals

4Chain rule

$\frac{d}{dx}\log(g(x)) = \frac{g'(x)}{g(x)}$ where $g(x) > 0$ ⁠. Basics: substitution (the chain rule backwards)

Solution

Check the integrand is continuous on $[0, 1]$

The numerator $x$ is a polynomial. The denominator satisfies $x^2 + 1 \ge 1$ for every real $x$ ⁠, so it is never 0. A quotient of continuous functions is continuous wherever the denominator is not 0, so $\frac{x}{x^2+1}$ is continuous on $[0, 1]$ (in fact on the whole real line).

Theorem 2 and the substitution rule both need a continuous integrand on the closed interval. Checking first rules out hidden trouble such as a zero denominator inside $[a, b]$ ⁠.

$x^2 \ge 0$ for every real $x$ ⁠, so $x^2 + 1 \ge 1$ ⁠. Lesson: the fundamental theorem
Substitute $t = x^2 + 1$ and change the limits

Differentiate: $\frac{dt}{dx} = 2x$ ⁠, so $dt = 2x\,dx$ ⁠, that is
$x\,dx = \frac{1}{2}\,dt.$
New limits. When $x = 0$ ⁠, $t = 0^2 + 1 = 1$ ⁠. When $x = 1$ ⁠, $t = 1^2 + 1 = 2$ ⁠.

Why it is valid on the whole interval. $t = x^2 + 1$ is a polynomial, so it has a continuous derivative $2x$ on $[0, 1]$ ⁠. Since $2x \ge 0$ there, $t$ increases steadily from 1 to 2, and every $t$ it passes through satisfies $t \ge 1 > 0$ ⁠, where $\frac{1}{t}$ is continuous.

The integrand is "a constant times $g'(x)$ over $g(x)$ " with $g(x) = x^2 + 1$ ⁠. Naming the denominator $t$ turns the whole fraction into $\frac{1}{2}\cdot\frac{dt}{t}$ ⁠. The limits must be changed to $t$ -values because after the substitution we integrate with respect to $t$ ⁠.

$\frac{d}{dx}(x^2 + 1) = 2x$ by the power rule. Basics: substitution
Integrate in $t$
$\begin{aligned}\int_0^1\frac{x\,dx}{x^2 + 1} &= \frac{1}{2}\int_1^2\frac{dt}{t} \\ &= \frac{1}{2}\Big[\log t\Big]_1^2 \\ &= \frac{1}{2}\left(\log 2 - \log 1\right) = \frac{1}{2}\log 2.\end{aligned}$
In decimals, $\frac{1}{2}\log 2 \approx \frac{0.6931472}{2} \approx 0.3465736$ ⁠.

On $[1, 2]$ ⁠, $t > 0$ ⁠, so $\log|t| = \log t$ ⁠. And $\log 1 = 0$ ⁠, so only the upper limit contributes.

Standard formula (xii), Section 7.2, then Theorem 2: F(upper) minus F(lower). Basics: logarithms and 1/x
Same result by staying in $x$

Returning to $x$ instead of changing the limits: $\frac{1}{2}\log t$ becomes
$F(x) = \frac{1}{2}\log(x^2 + 1).$
Check: differentiate. By the chain rule (outer $\frac{1}{2}\log u$ ⁠, inner $u = x^2 + 1$ ⁠),
$F'(x) = \frac{1}{2}\cdot\frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1},$
the integrand. Then $F(1) - F(0) = \frac{1}{2}\log 2 - \frac{1}{2}\log 1 = \frac{1}{2}\log 2$ ⁠, as before.

Steps 1 to 3 of Section 7.9 (go back to $x$ ⁠) and the quick method of its Note (new limits) must give the same number. Doing both catches a wrongly changed limit.

No modulus is needed: $x^2 + 1 > 0$ for every $x$ ⁠. Basics: the chain rule backwards
The region

$y = \frac{x}{x^2+1}$ rises from 0 to its highest value $\frac{1}{2}$ at $x = 1$ ⁠. The shaded area from 0 to 1 is $\frac{1}{2}\log 2 \approx 0.347$ ⁠.

On $[0, 1]$ the integrand is $\ge 0$ ⁠, so the integral is the shaded area. A quick bound: for $0 \le x \le 1$ ⁠, $1 \le x^2 + 1 \le 2$ ⁠, so $\frac{x}{2} \le \frac{x}{x^2+1} \le x$ ⁠. The area therefore lies between $\int_0^1\frac{x}{2}dx = 0.25$ and $\int_0^1 x\,dx = 0.5$ ⁠, and 0.347 does.

For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

Result

$\int_0^1\frac{x}{x^2 + 1}\,dx = \frac{1}{2}\log 2 \approx 0.3465736.$ (Textbook answer: $\frac{1}{2}\log 2$ ⁠.)

Check it with numbers

A numerical integration of $\frac{x}{x^2+1}$ over $[0, 1]$ gives 0.3465736, and $\frac{1}{2}\log 2 \approx \frac{0.6931472}{2} = 0.3465736$ ⁠. Halfway, $F(0.5) - F(0) = \frac{1}{2}\log 1.25 \approx 0.1115718$ ⁠: less than a third of the total, because the curve is low near $x = 0$ and highest near $x = 1$ ⁠.

Common mistakes

Keeping the old limits. After $t = x^2 + 1$ the limits are 1 and 2, not 0 and 1. Using 0 and 1 in $\frac{1}{2}\log t$ would even give $\log 0$ ⁠, which does not exist.
Forgetting the $\frac{1}{2}$ ⁠. $dt = 2x\,dx$ ⁠, so $x\,dx$ is only half of $dt$ ⁠. Without the $\frac{1}{2}$ the answer would be $\log 2$ ⁠, twice too big.
Writing $\int\frac{x}{x^2+1}dx = \frac{x^2/2}{x^3/3 + x}$ ⁠. There is no quotient rule for integrals; the quotient is handled by substitution.

What to take away

When the numerator is a constant multiple of the derivative of the denominator, substitute $t =$ denominator: the answer is a logarithm.
Change the limits together with the variable: $t = g(a)$ to $t = g(b)$ ⁠.
Check that the substitution and $\frac{1}{t}$ behave on the whole interval: here $t$ runs through $[1, 2]$ ⁠, away from 0.

Every fact used, and where it comes from

Fact	Used in	Source
A quotient of continuous functions is continuous where the denominator is not 0	Step 1	Chapter 5, Section 5.2.1 (algebra of continuous functions)
Substitution with change of limits	Step 2	Section 7.9 and its Note (Lesson)
$\int\frac{dt}{t} = \log\|t\| + C$	Step 3	Standard formula (xii), Section 7.2 (Basics)
$\int_a^b f = F(b) - F(a)$	Steps 3 and 4	Theorem 2, Section 7.8.3 (Lesson)
Chain rule for $\log(g(x))$	Step 4	Chapter 5, Sections 5.3.1 and 5.4

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Exercise 7.9 · Q1: ∫01xx2+1 dx\int_0^1\frac{x}{x^2+1}\,dx∫01​x2+1x​dx

Basics you need

Solution

Check the integrand is continuous on [0,1][0, 1][0,1]

Substitute t=x2+1t = x^2 + 1t=x2+1 and change the limits

Integrate in ttt

Same result by staying in xxx

The region

Every fact used, and where it comes from

Exercise 7.9 · Q1: $\int_0^1\frac{x}{x^2+1}\,dx$

Check the integrand is continuous on $[0, 1]$

Substitute $t = x^2 + 1$ and change the limits

Integrate in $t$

Same result by staying in $x$