Biết I=∫224x2−1xx2−4dx=π2a+b3+ca,b,c∈ℤ*. Khi đó, P=−α+b+c bằng

Cách 1. Ta có

=∫224x2−1xx2−4dx=∫224xx2−4−1xx2−4dx=∫224xx2−4dx−∫2241xx2−4dx=∫224xx2−4dx−π24

Đặt t=x2−4⇒t2=x2−4⇒tdt=xdx

Với x=22⇒t=2,x=4⇒t=23

⇒I=∫223dt−π24=t223−π24=23−2−π24⇒a=−12,b=2,c=−2⇒P=12

Cách 2. Đặt x=2cost. Ta có dx=−2sintcos2tdt24=2cosπ3222=2cosπ3

⇒I=∫π4π34cos2t−12cost4cos2t−4−2sintdtcos2t=12∫π4π3cos2t−4cos2tdt=π24−23+2

⇒a=−12,b=2,c=−2⇒P=12

Chọn D