Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
The indefinite integral expressed as an infinite series is;
[tex]= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C[/tex]
How to find indefinite integral?
We will first have to look for the Maclaurin series of arctan(x).
We'll recall that from online tables of integral, this Maclaurin series of arctan(x) will have the general formula;
[tex]arctan(x) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{x^{2n + 1} }{2n + 1}[/tex]
When we apply that general Maclaurin series of arctan(x) to our question of arctan(x²), we have the expression as;
[tex]arctan(x^{2} ) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x^2)^{2n + 1} }{2n + 1}[/tex]
⇒ [tex]= \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1}[/tex]
We now integrate the expression that we got above in the following manner to get;
[tex]\int\limitsarctan(x^{2} ) = \int\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1} dx[/tex]
⇒ [tex]= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C[/tex]
Thus, that expression gives us the indefinite integral of arctan(x²) as an infinite series.
Read more about the indefinite integral at; https://brainly.com/question/12231722
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
