Chapter 7 · Integrals

Exercise 7.6 · Q8: $\int x\tan^{-1}x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

8. $x\tan^{-1}x$

Exam answer

Let $I = \displaystyle\int x\tan^{-1}x\,dx$ ⁠.

Taking $\tan^{-1}x$ as the first function, by parts:

\begin{aligned} I &= \tan^{-1}x\cdot\frac{x^2}{2} \\ &\quad - \int\frac{1}{1 + x^2}\cdot\frac{x^2}{2}\,dx \\ &= \frac{x^2}{2}\tan^{-1}x \\ &\quad - \frac12\int\left(1 - \frac{1}{1 + x^2}\right)dx \\ &= \frac{x^2}{2}\tan^{-1}x \\ &\quad - \frac12\left(x - \tan^{-1}x\right) + C \end{aligned}

$\therefore\ I = \dfrac{x^2}{2}\tan^{-1}x - \dfrac x2$ $+ \dfrac12\tan^{-1}x + C$ ⁠.

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

The one idea this problem is built on

Differentiate the inverse tangent (it becomes the algebraic $\frac{1}{1+x^2}$ ⁠) and integrate $x$ ⁠. The integral that is left, $\int\frac{x^2}{1+x^2}\,dx$ ⁠, is a fraction whose top has the same degree as the bottom; dividing gives $1 - \frac{1}{1+x^2}$ ⁠, and both pieces are standard.

Given, to find, and the plan

Given: The function $x\tan^{-1}x$ ⁠, defined and continuous for every real $x$ ( $\tan^{-1}$ is defined on all of $\mathbf R$ ⁠).
To find: $\int x\tan^{-1}x\,dx$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ ⁠. (2) Find $\frac{du}{dx}$ and $v$ ⁠. (3) Apply the formula. (4) Split $\frac{x^2}{1+x^2}$ and integrate. (5) Differentiate back.

Basics you need

1Integration by parts

Section 7.6, equation (1):

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

with $u$ the first function (differentiated) and $\frac{dv}{dx}$ the second (integrated to give $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

Remark (iii) after Example 17: an inverse trigonometric function is taken as the first function. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): I before A.

3The inverse tangent

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ ⁠, so $\int\frac{dx}{1+x^2} = \tan^{-1}x + C$ (standard formula (x)). Basics: trigonometric identities

4Splitting a fraction

When the top and bottom have the same degree, divide first: $\frac{x^2}{1+x^2} = \frac{(1+x^2) - 1}{1+x^2}$ ⁠. Basics: splitting fractions

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= \tan^{-1}x \quad (\text{first}), \\ \frac{dv}{dx} &= x \quad (\text{second}). \end{aligned}$
ILATE: Inverse trigonometric before Algebraic. We have no formula for $\int\tan^{-1}x\,dx$ at this point, but its derivative $\frac{1}{1+x^2}$ is a simple algebraic fraction. So $\tan^{-1}x$ is the function to differentiate.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = \frac{1}{1+x^2}, \qquad v = \int x\,dx = \frac{x^2}{2}.$
Derivative of the inverse tangent (valid for every real $x$ ⁠); $v$ by the power rule, with no constant (Remark (ii) after Example 17).

Standard formulae (x) and (i), Section 7.2. Basics: powers of x
Apply the formula
$\begin{aligned} \int x\tan^{-1}x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= \frac{x^2}{2}\tan^{-1}x - \frac12\int\frac{x^2}{1+x^2}\,dx. \end{aligned}$
The inverse tangent has left the remaining integral, which is now a rational function of $x$ ⁠.

Section 7.6, equation (1). Lesson: integration by parts
Split the fraction and integrate
$\begin{aligned} \frac{x^2}{1+x^2} &= \frac{(1+x^2) - 1}{1+x^2} = 1 - \frac{1}{1+x^2}, \\ \int\frac{x^2}{1+x^2}\,dx &= x - \tan^{-1}x. \end{aligned}$
So
$\begin{aligned} \int x\tan^{-1}x\,dx &= \frac{x^2}{2}\tan^{-1}x - \frac12\left(x - \tan^{-1}x\right) + C \\ &= \frac{x^2}{2}\tan^{-1}x - \frac x2 + \frac12\tan^{-1}x + C. \end{aligned}$
Adding and subtracting 1 in the numerator makes the top equal to the bottom plus a constant, so the fraction splits into 1 and a standard fraction. Then multiply by $-\frac12$ ⁠: the $\tan^{-1}x$ term gets a plus sign.

Splitting fractions; standard formulae (i) and (x). Basics: splitting fractions
Check: differentiate the answer

Product rule on the first term:
$\begin{aligned} &\frac{d}{dx}\Big(\frac{x^2}{2}\tan^{-1}x - \frac x2 + \frac12\tan^{-1}x\Big) \\ &= x\tan^{-1}x + \frac{x^2}{2(1+x^2)} - \frac12 + \frac{1}{2(1+x^2)} \\ &= x\tan^{-1}x + \frac{x^2 + 1}{2(1+x^2)} - \frac12 \\ &= x\tan^{-1}x. \end{aligned}$
The two fractions add to $\frac12$ ⁠, which cancels the $-\frac12$ ⁠. Only the integrand is left, for every real $x$ ⁠.

Product rule; $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ ⁠. Basics: integration undoes differentiation

Result

\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x - \frac{x}{2} + \frac12\tan^{-1}x + C,

valid for every real $x$ ⁠. It can also be written $\frac{x^2+1}{2}\tan^{-1}x - \frac x2 + C$ ⁠. (Textbook answer: $\frac{x^2}{2}\tan^{-1}x - \frac x2 + \frac12\tan^{-1}x + C$ ⁠, the same.)

Check it with numbers

Call the answer $F(x)$ ⁠. On $[0, 1]$ ⁠, with $\tan^{-1}1 = \frac\pi4$ ⁠:

\begin{aligned} F(1) - F(0) &= \frac12\cdot\frac\pi4 - \frac12 + \frac12\cdot\frac\pi4 - 0 \\ &= \frac\pi4 - \frac12 = 0.2853981634\ldots \end{aligned}

Numerical integration of $x\tan^{-1}x$ over $[0, 1]$ gives $0.2853981634\ldots$ ⁠, the same.

Common mistakes

Taking $u = x$ ⁠. Then $v = \int\tan^{-1}x\,dx$ is needed, which we do not have.
Integrating $\frac{x^2}{1+x^2}$ as a logarithm. The top is not the derivative of the bottom; divide first.
Sign of the last term. $-\frac12(x - \tan^{-1}x) = -\frac x2 + \frac12\tan^{-1}x$ ⁠.

What to take away

Inverse trigonometric function times a power of $x$ ⁠: the inverse function is $u$ ⁠.
A fraction with top degree at least the bottom degree: divide (or add and subtract) before integrating.
Collecting terms, $\frac{x^2+1}{2}\tan^{-1}x - \frac x2$ is a neat equivalent form.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Step 3	Section 7.6, equation (1) (Lesson: integration by parts)
Inverse trigonometric function as the first function (ILATE)	Step 1	Remark (iii) after Example 17, Section 7.6
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ ⁠, $\int\frac{dx}{1+x^2} = \tan^{-1}x + C$	Steps 2, 4, 5	Standard formula (x), Section 7.2
Power rule	Steps 2, 4	Standard formula (i), Section 7.2
Splitting $\frac{x^2}{1+x^2}$	Step 4	Algebra of fractions (as in Section 7.5, improper fractions)
Product rule	Step 5	Chapter 5, Section 5.3

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Exercise 7.6 · Q8: ∫xtan⁡−1x dx\int x\tan^{-1}x\,dx∫xtan−1xdx

Basics you need

Solution

Name uuu and dvdx\frac{dv}{dx}dxdv​

Find dudx\frac{du}{dx}dxdu​ and vvv

Apply the formula

Split the fraction and integrate

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q8: $\int x\tan^{-1}x\,dx$

Name $u$ and $\frac{dv}{dx}$

Find $\frac{du}{dx}$ and $v$