|
Ta có 1(x−1)√x2−4x+3=x−3(x−1)(x−3)√x2−4x+3
1(x−1)√x2−4x+3=(x2−4x+3)−(x−2)(x−3)√x2−4x+3x2−4x+3
1(x−1)√x2−4x+3=√x2−4x+3−(x−2)(x−3)√x2−4x+3x2−4x+3
1(x−1)√x2−4x+3=√x2−4x+3(x−3)′−(√x2−4x+3)′(x−3)x2−4x+3
1(x−1)√x2−4x+3=(x−3√x2−4x+3)′
Suy ra
4∫2+√21(x−1)√x2−4x+3dx=[x−3√x2−4x+3]42+√2=1−√2+1√3
|