Ta có
$I = \int_0^1 \dfrac{x^{2n-4}xdx}{(x^2+1)^n}$
$= \dfrac{1}{2} \int_0^1 \dfrac{(x^2)^{n-2}d(x^2)}{(x^2+1)^n}$
Đặt $t = x^2$. Khi đó ta có
$2I = \int_0^1\dfrac{t^{n-2}dt}{(t+1)^n}$
$= \int_0^1 \left( \dfrac{t}{t+1} \right)^n . \dfrac{1}{t^2} dt$
$= -\int_0^1 \left( 1 - \dfrac{1}{t+1} \right) d\left( \dfrac{1}{t+1} \right)$
Đặt $u = \dfrac{1}{t+1}$. KHi đó $u$ sẽ chạy từ $\dfrac{1}{2}$ đến 1.
$-2I = (-1)^n \int_{\frac{1}{2}}^1 (u-1)^n dx$
$= (-1)^n \dfrac{(u-1)^{n+1}}{n+1}\bigg\vert_{\frac{1}{2}}^1$
$= (-1)^n \left[ \left( -\dfrac{1}{2} \right)^{n+1} . \dfrac{1}{n+1} - 0\right]$
$= (-1)^n (-1)^{n+1} . \left( \dfrac{1}{2} \right)^{n+1}$
$= -\left( \dfrac{1}{2} \right)^{n+1}$
Suy ra
$I = \left( \dfrac{1}{2} \right)^{n+2}$.
