Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
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
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
