Smelling, seeing, hearing, tasting and touching.
Answer:
yes that's a very cute fox
Explanation:
:)
Answer:
Why has GATT taken longer than expected?
The indefinite integral expressed as an infinite series is;
![= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C](https://tex.z-dn.net/?f=%3D%20%20%28%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7B1%20%7D%7B2n%20%2B%201%7D%20%2A%20%5Cfrac%7B%28x%29%5E%7B4n%20%2B%203%7D%7D%7B4n%20%2B%203%7D%29%20%2B%20C)
<h3>How to find indefinite integral?</h3>
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;
![arctan(x) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{x^{2n + 1} }{2n + 1}](https://tex.z-dn.net/?f=arctan%28x%29%20%3D%20%20%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7Bx%5E%7B2n%20%2B%201%7D%20%7D%7B2n%20%2B%201%7D)
When we apply that general Maclaurin series of arctan(x) to our question of arctan(x²), we have the expression as;
![arctan(x^{2} ) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x^2)^{2n + 1} }{2n + 1}](https://tex.z-dn.net/?f=arctan%28x%5E%7B2%7D%20%29%20%3D%20%20%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7B%28x%5E2%29%5E%7B2n%20%2B%201%7D%20%7D%7B2n%20%2B%201%7D)
⇒ ![= \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1}](https://tex.z-dn.net/?f=%3D%20%20%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7B%28x%29%5E%7B4n%20%2B%202%7D%20%7D%7B2n%20%2B%201%7D)
We now integrate the expression that we got above in the following manner to get;
![\int\limitsarctan(x^{2} ) = \int\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1} dx](https://tex.z-dn.net/?f=%5Cint%5Climitsarctan%28x%5E%7B2%7D%20%29%20%3D%20%20%5Cint%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7B%28x%29%5E%7B4n%20%2B%202%7D%20%7D%7B2n%20%2B%201%7D%20dx)
⇒ ![= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C](https://tex.z-dn.net/?f=%3D%20%20%28%5CSigma%5E%7B%5Cinfty%7D%20_%7Bn%20%3D%200%7D%20%28-1%29%5E%7Bn%7D%20%5Cfrac%7B1%20%7D%7B2n%20%2B%201%7D%20%2A%20%5Cfrac%7B%28x%29%5E%7B4n%20%2B%203%7D%7D%7B4n%20%2B%203%7D%29%20%2B%20C)
Thus, that expression gives us the indefinite integral of arctan(x²) as an infinite series.
Read more about the indefinite integral at; brainly.com/question/12231722