MHB Evaluating a rational function with contour integration

[Amad27]

Hello, I am looking to evaluate:

$$I = \int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2} dx$$

I will use a rectangular contour.

The image looked weird here so the upload of the image is here:

http://i.stack.imgur.com/W4BfA.jpg

$R$ is more like the radius of the small semi circle, we have to let $R \to 1$ in the end.

The pole $z = i$ is inside, so by the residue theorem,

$$\oint_{C} f(z) dz = -4\pi$$

I have never used a square contour before, so will someone help me out?

Thanks.

 

[Prove It]

Hello, I am looking to evaluate:

$$I = \int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2} dx$$

I will use a rectangular contour.

The image looked weird here so the upload of the image is here:

http://i.stack.imgur.com/W4BfA.jpg

$R$ is more like the radius of the small semi circle, we have to let $R \to 1$ in the end.

The pole $z = i$ is inside, so by the residue theorem,

$$\oint_{C} f(z) dz = -4\pi$$

I have never used a square contour before, so will someone help me out?

Thanks.

$\displaystyle \begin{align*} \frac{x^4 \, \left( 1 - x \right) ^4 }{ x^2 + 1 } &= \frac{ x^4 \left( 1 - 4x + 6x^2 - 4x^3 + x^4 \right) }{x^2 +1} \\ &= \frac{x^8 - 4x^7 + 6x^6 - 4x^5 + x^4}{x^2 + 1} \\ &= x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{x^2 + 1} \end{align*}$

this should be very easy to integrate now...