Integrate the following using infinite #\bb ext(series)# ?

Question

(a) #\int(\arctan(4x))/(x^-7)dx#

Christopher Agresta · Accepted Answer

#int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/((2n+1)(2n+9) )x^(2n+9) +C#

Start from:

#arctan(4x) = int_0^(4x) dt/(1+t^2)#

Expand now the integrand using the geometric series:

#arctan(4x) = int_0^(4x) sum_(n=0)^oo (-1)^nt^(2n)dt#

for #abs (4x) < 1# the series is absolutely convergent and we can integrate term by term:

#arctan(4x) = sum_(n=0)^oo (-1)^nint_0^(4x) t^(2n)dt#

#arctan(4x) = sum_(n=0)^oo (-1)^n (4x)^(2n+1)/(2n+1)#

#arctan(4x) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+1)#

Dividing by #x^(-7)# means multiplying by #x^7# and we can do it again term by term:

#arctan(4x)/x^(-7) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+8)#

and integrating again term by term:

#int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)int x^(2n+8) dx#

#int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/((2n+1)(2n+9) )x^(2n+9) +C#

Isaiah Allen · Answer

undefined To integrate using an infinite series, follow these general steps:

1. Express the function as a power series if possible.
2. Determine the interval of convergence for the power series.
3. Integrate the power series term by term within the interval of convergence.

If you have a specific function or expression you'd like to integrate using an infinite series, please provide that information so I can assist you further.

Integrate the following using infinite #\bb\text(series)# ?

(a) #\int(\arctan(4x))/(x^-7)dx#