Question

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

`\int(4x^3)/(1+x^8)dx`

(If there are trig substitution steps, please explain if possible -- I'm pretty bad at those)

Answer 1

`int \ (4x^3)/(1+x^8) \ dx = 4x - (1)/(3)x^12 + (1)/(5)x^20 - (1)/(7)x^28 + ... + C`

Explanation:

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.

Answer 2

See below.

Explanation:

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`