south_korea_ln

science

[Daily puzzle] Integral

This is as far as I got.  Maybe someone else can take it the rest of the way

$$
u = x^4
$$

$$
du = 4x^3 dx
$$

$$
\int_{0}^{1} \frac{4x^3 \left(1+x^{4(2006)}\right)}{(1+x^4)^{2008}} dx = \int_{0}^{1} \frac{1+u^{2006}}{(1+u)^{2008}} du
$$

$$
= \int_{0}^{1} \frac{1}{(1+u)^{2008}} du + \int_{0}^{1} \frac{u^{2006}}{(1+u)^{2008}} du
$$

The left term can be easily calculated.

For the right term, let

$$
F(k) = \int_{0}^{1} \frac{u^{k}}{(1+u)^{k+2}} du 
$$

Using integration by parts, you can show that:

$$
F(k) = \left. -\frac{u^{k}}{(k+1)(1+u)^{k+1}} \right|_{0}^{1} + \frac{k}{k+1} F(k-1)
$$

Starting at $$F(0)$$ which is easy to calculate, you can use the above algorithm to eventually calculate $$F(2006)$$ to find the answer.