Problems on Techniques of Integration

The Arctangent function is one of these functions whose derivative is very nice (rational function). So it is a good idea to use the integration by parts technique in which $\arctan(x)$ will be differentiated.

Set

$$\left\{\begin{array}{lll} u &=& \arctan(x)\\ dv &=& x dx\;. \end{array}\right.$$

Then

$$\left\{\begin{array}{lll} du &=&\displaystyle \frac{1}{x^2 + 1} dx\\ v &=& \displaystyle \frac{x^2}{2}\;. \end{array}\right.$$

Since

$$\int u dv = u v - \int v du\;,$$

we get

$$\int x \arctan(x) dx = \frac{x^2}{2} \arctan(x) - \int \frac{... ...x^2}{2} \arctan(x) - \frac{1}{2}\int \frac{x^2}{x^2 + 1}dx\cdot$$

In order to integrate the function $$\frac{x^2}{x^2 + 1}$$, we will need the technique of integrating rational functions. But in this case, we may use the identity

$$\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$$

which implies

$$\int \frac{x^2}{x^2 + 1}dx = \int \left(1 - \frac{1}{x^2 + 1}\right)dx = x - \arctan(x)\cdot$$

So

$$\int \arctan(x) dx = \frac{x^2}{2} \arctan(x)- \frac{x}{2} + \frac{1}{2}\arctan(x) + C\;.$$

It is a common mistake to forget the constant $C$.

Short Answer.

Detailed Answer.

If you prefer to jump to the next problem, click on Next Problem below.

[Next Problem] [Matrix Algebra] [Trigonometry] [Calculus] [Geometry] [Algebra] [Differential Equations] [Complex Variables]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Copyright © 1999-2025 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour