Question

Evaluate the indefinite integral as an infinite series. arctan(x2) dx

Answer 1

The indefinite integral expressed as an infinite series  is;

$= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C$

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;

$arctan(x) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{x^{2n + 1} }{2n + 1}$

When we apply that general Maclaurin series of arctan(x) to our question of arctan(x²), we have the expression as;

$arctan(x^{2} ) = \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x^2)^{2n + 1} }{2n + 1}$

⇒ $= \Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1}$

We now integrate the expression that we got above in the following manner to get;

$\int\limitsarctan(x^{2} ) = \int\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{(x)^{4n + 2} }{2n + 1} dx$

⇒ $= (\Sigma^{\infty} _{n = 0} (-1)^{n} \frac{1 }{2n + 1} * \frac{(x)^{4n + 3}}{4n + 3}) + C$

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