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

## $\setminus \int \frac{4 {x}^{3}}{1 + {x}^{8}} \mathrm{dx}$ (If there are trig substitution steps, please explain if possible -- I'm pretty bad at those)

May 12, 2018

$\int \setminus \frac{4 {x}^{3}}{1 + {x}^{8}} \setminus \mathrm{dx} = 4 x - \frac{1}{3} {x}^{12} + \frac{1}{5} {x}^{20} - \frac{1}{7} {x}^{28} + \ldots + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{4 {x}^{3}}{1 + {x}^{8}} \setminus \mathrm{dx}$

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

$I = \int \setminus 4 {x}^{3} {\left(1 + {x}^{8}\right)}^{- 1} \setminus \mathrm{dx}$

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 + ... }

$\setminus \setminus = \int \setminus \left(4 {x}^{3}\right) \left\{1 - {x}^{8} + {x}^{16} - {x}^{24} + \ldots\right\}$

$\setminus \setminus = \int \setminus 4 - 4 {x}^{11} + 4 {x}^{19} - 4 {x}^{27} + \ldots$

$\setminus \setminus = 4 x - \frac{4}{12} {x}^{12} + \frac{4}{20} {x}^{20} - \frac{4}{28} {x}^{28} + \ldots + C$

$\setminus \setminus = 4 x - \frac{1}{3} {x}^{12} + \frac{1}{5} {x}^{20} - \frac{1}{7} {x}^{28} + \ldots + C$

Note:

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

$u = {x}^{4} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 4 {x}^{3}$

And if we substitute into the integral, we gtre:

$I = \int \setminus \frac{1}{1 + {u}^{2}} \setminus \mathrm{du}$

Which is a standard integral, so we get:

$I = \arctan u + C$

And restoration of the substitution gives:

$I = \arctan \left({x}^{4}\right) + C$

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

May 12, 2018

See below.

#### Explanation:

Here's a solution without the Binomial Theorem, and rather just using the Power Series $\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$ :

$\int 4 {x}^{3} \cdot \frac{1}{1 + {x}^{8}} \mathrm{dx} = \int 4 {x}^{3} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{8 n} \mathrm{dx}$

As

$\frac{1}{1 + {x}^{8}} = \frac{1}{1 - \left(- {x}^{8}\right)} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{8}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{8 n}$

Multiply in ${x}^{3} :$

$= \int 4 {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{8 n + 3} \mathrm{dx}$

$= 4 {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \int {x}^{8 n + 3} \mathrm{dx}$

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

$= C + {\sum}_{n = 0}^{\infty} 4 {\left(- 1\right)}^{n} {x}^{8 n + 4} / \left(8 n + 4\right)$

$= C + {\sum}_{n = 0}^{\infty} \cancel{4} {\left(- 1\right)}^{n} {x}^{4 \left(2 n + 1\right)} / \left(\cancel{4} \left(2 n + 1\right)\right)$

$= C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left({x}^{4}\right)}^{2 n + 1} / \left(2 n + 1\right)$

Now, recalling that $\arctan x = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right) ,$

$C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left({x}^{4}\right)}^{2 n + 1} / \left(2 n + 1\right) = \arctan \left({x}^{4}\right) + C$