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

my discord name is kinglilcliff12#7776

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)