Evaluate the indefinite integral as a power series.

∫ tan-1 (x) / x dx

[infinity]

f (x) = C + Σ

n = 0

What is the radius of convergence R?

- The integral of arctan (x) / x =

C + Σ [ (-1) ^n. x^ (2n+1) ] / (2n+1) ². (From n = 0 to infinity).

- The radius of convergence is R = 1/x

Step-by-step explanation:

First note that

tan^ (-1) x = arctan (x)

And

arctan (x) = Σ [ (-1) ^n. x^ (2n+1) ] / (2n+1). From n = 0 to infinity

arctan (x) / x = Σ [ (-1) ^n. x^ (2n) ] / (2n+1). From n = 0 to infinity

∫arctan (x) / x dx = ∫{Σ [ (-1) ^n. x^ (2n) ] / (2n+1). From n = 0 to infinity}dx

= ∫{Σ [ (-1) ^n] / (2n+1). From n = 0 to infinity}.∫x^ (2n) dx

= {Σ [ (-1) ^n] / (2n+1). From n = 0 to infinity}. x^ (2n+1) / (2n+1) + C

= C + Σ [ (-1) ^n. x^ (2n+1) ] / (2n+1) ². (From n = 0 to infinity).

To obtain the radius of convergence, we apply the ration test

R = Limit as n approaches infinity |a_n/a_ (n+1) |

a_n = (-1) ^n. x^ (2n+1) ] / (2n+1) ²

a_ (n+1) = (-1) ^ (n+1). x^ (2 (n+1) + 1) ] / (2 (n+1) + 1) ²

|a_n/a_ (n+1) | = (2 (n+1) + 1) ² / (2n+1) ². x

= (1/x) [1 + 2 / (2n+1) ]

R = Limit as n approaches infinity |a_n/a_ (n+1) |

= R = Limit as n approaches infinity (1/x) [1 + 2 / (2n+1) ]

R = 1/x