$=\int_0^1 (t^4-t^2+1-\dfrac{1}{t^2+1})dt=(\dfrac{1}{5}t^5-\dfrac{1}{3}t^3+t)\bigg|_0^1 -\int_0^1 \dfrac{1}{t^2+1}dt$
Đặt $t=\tan u\Rightarrow dt =\dfrac{du}{\cos^2 u}$
$\int_0^1 \dfrac{1}{t^2+1}dt =\int \dfrac{1}{1+\tan^2 u} .\dfrac{du}{\cos^2 u} =\int du= u \bigg|_0^{\frac{\pi}{4}}=....$
Tự hoàn thiện