Ah i was trying to hint at the fact one had to do another substitution. You had already done one, i was trying to tell you to do another one. Instead of switching to integration by parts. Maybe not the best hint ;)

SimpleStacker

science

[Daily puzzle] Integral

south_korea_ln

First, break the integral into two parts as
$$
\int_0^1 \frac{4x^3}{(1 + x^4)^{2008}} \, dx + \int_0^1 \frac{4x^3 \cdot x^{4(2006)}}{(1 + x^4)^{2008}} \, dx.
$$

Using the $$u = 1 + x^4$$ substitution, the first integral becomes

$$
\int_1^2 \frac{1}{u^{2008}} \, du = \frac{1}{2007}(1 - 2^{-2007}),
$$
while the second becomes
$$
\int_1^2 \frac{(u - 1)^{2006}}{u^{2008}} \, du = \int_1^2 \frac{(1 - 1/u)^{2006}}{u^2} \, du.
$$

With the additional substitution $$v = 1 - 1/u$$, the final integral above becomes

$$
\int_0^{1/2} v^{2006} \, dv = \frac{1}{2007}(2^{-2007})
$$
so that the original integral is equal to
$$
\frac{1}{2007}(1 - 2^{-2007}) + \frac{1}{2007}(2^{-2007}) = \frac{1}{2007}.
$$
