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

We seek:

# I = int \ (4x^3)/(1+x^8) \ dx #

in the form of a Power Series. We can write the integral in the form:

# I = int \ 4x^3(1+x^8)^(-1) \ dx #

And we can utilise the Binomial Theorem:

# (1+x)^n = 1 + nx + (n(n-1))/(2!)x^2 + (n(n-1)(n-2))/(3!)x^3 + ...#

So then we can write:

# I = int \ (4x^3){1+(-1)(x^8) + ((-1)(-2))/(2!)(x^8)^2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3))/(3!)(x^8)^3 + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3)(-4))/(4!)(x^8)^4 + ... }#

# \ \ = int \ (4x^3){1-x^8 + x^16 - x^24 + ... } #

# \ \ = int \ 4 - 4x^11 + 4x^19 - 4x^27 + ... #

# \ \ = 4x - (4)/(12)x^12 + (4)/(20)x^20 - (4)/(28)x^28 + ... + C#

# \ \ = 4x - (1)/(3)x^12 + (1)/(5)x^20 - (1)/(7)x^28 + ... + C #

Note:

In this particular case we can evaluate the integration by direct substitution, Let

# u = x^4 => (du)/dx = 4x^3 #

And if we substitute into the integral, we gtre:

# I = int \ 1/(1+u^2) \ du #

Which is a standard integral, so we get:

# I = arctanu+C #

And restoration of the substitution gives:

# I = arctan(x^4) + C#

And it can readily be verified that this solution has the above Power Series expansion.

Here's a solution without the Binomial Theorem, and rather just using the Power Series #1/(1-x)=sum_(n=0)^oox^n# :

#int4x^3*1/(1+x^8)dx=int4x^3sum_(n=0)^oo(-1)^nx^(8n)dx#

As

#1/(1+x^8)=1/(1-(-x^8))=sum_(n=0)^oo(-x^8)^n=sum_(n=0)^oo(-1)^nx^(8n)#

Multiply in #x^3:#

#=int4sum_(n=0)^oo(-1)^nx^(8n+3)dx#

#=4sum_(n=0)^oo(-1)^nintx^(8n+3)dx#

Integrate term-by-term (don't forget constant of integration):

#=C+sum_(n=0)^oo4(-1)^nx^(8n+4)/(8n+4)#

#=C+sum_(n=0)^oocancel4(-1)^nx^(4(2n+1))/(cancel4(2n+1))#

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

Now, recalling that #arctanx=sum_(n=0)^oo(-1)^nx^(2n+1)/(2n+1),#

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