This problem aims to familiarize us with the **power series of an indefinite integral**.

This question requires the understanding of **fundamental** **calculus,** which includes **indefinite integrals, Power series, **and **the radius of convergence.**

Now,** Indefinite integrals** are mostly normal integrals but are expressed without **higher **and **lower limits** on the integrand, the expression $\int f(x)$ is employed to represent the **function** as an **antiderivative** of the function.

Whereas a **power series** is an indefinite series of the form $ \sum_{n= 0}^{ \infty } a_n (x – c)^{n} $ where $a_n$ symbolizes the **coefficient** of the $n^{th}$ duration and $c$ represents a **constant.** Such **power series** are helpful in mathematical analysis, and are transformed into **Taylor series** to solve infinitely **differentiable** expressions.

## Expert Answer

If we expand the **expression** $tan^{-1}x$ into an **indefinite** summation, we get something as follows:

\[x – \dfrac{x^3}{3} + \dfrac{x^5}{5} – \dfrac{x^7}{7} + \dfrac{x^9}{9} \space ….. \]

The given **integral** can be written as a **power series:**

\[\int \dfrac{tan^{-1}x}{x} dx = \int \dfrac{1}{x} \left( x – \dfrac{x^3}{3} + \dfrac{x^5}{5} – \dfrac{x^7}{7} + \dfrac{x^9}{9} \space …. \right) dx\]

\[= \int \left( 1 – \dfrac{x^2}{3} + \dfrac{x^4}{5} – \dfrac{x^6}{7} + \dfrac{x^8}{9} \space …. \right) dx\]

By solving the **integral:**

\[=x – \dfrac{x^3}{9} + \dfrac{x^5}{25} – \dfrac{x^7}{49} + \dfrac{x^9}{81} \space ….\]

This above **sequence** can be written in the form of:

\[=\sum_{n= 1}^{ \infty } \dfrac{ (-1) ^ {n-1} x^{2n -1}} {( 2n -1 )^2 }\]

Which is the required **power series.**

The **radius** of **convergence** is given as:

\[R = lim_{n \rightarrow \infty } \left| \dfrac{a_n}{a_{n+1}} \right|\]

Here, we have:

\[a_n = \dfrac{ (-1) ^ {n-1} } {( 2n -1 )^2 }\]

\[a_{n+1} = \dfrac{ (-1) ^ {n} } {( 2n +1 )^2 }\]

**Therefore**:

\[R = lim_{n \rightarrow \infty } \left| \dfrac{( -1)^{n-1}}{ (2n -1 )^2 } \times \dfrac { (2n +1 )^2 } {( -1)^{n}} \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{(2n+1)^{2}}{ (2n -1 )^2 } \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{4n^2 \left( 1 + \dfrac{1}{2n} \right)^2 }{ 4n^2 \left( 1 – \dfrac{1}{2n} \right)^2 } \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{ \left( 1 + \dfrac{1}{2n} \right)^2 }{ \left( 1 – \dfrac{1}{2n} \right)^2 } \right|\]

Therefore, the **radius** of **convergence** is $R = 1$.

## Numerical Result

**Indefinite integral** as a **power series** is $ \sum_{n= 1}^{ \infty } \dfrac{ (-1) ^ {n-1} x^{2n -1}} {( 2n -1 )^2 } $.

**Radius** of convergence is $ R =1 $.

## Example

Using the **Power Series,** evaluate the given integral $ \int \dfrac{x}{1+x^3} dx $.

The given **integral** can be written as a **power** series as follows:

\[ = \sum_{n= 0}^{ \infty } (-1) ^ {n} x^{3n +1} \]

The series **converges** when $|-x^3| < 1$ or $|x| < 1$ , so for this particular **power series** $R = 1$.

Now, we **integrate:**

\[ \int \dfrac{x}{1+x^3} dx = \int \sum_{n= 0}^{ \infty } (-1) ^ {n} x^{3n +1} dt \]

**Indefinite integral** as a power series comes out to be:

\[ = C + \sum_{n= 0}^{ \infty } (-1) ^ {n} \dfrac{ x^{3n +2}}{( 3n +2 ) } \]