a)I=7/9∫2/3dx√(x−1)(x−13)Đặt(x−1)t=√(x−1)(x−13)⇔(x−13)t2=(x−1)⇔x=t2−33t−3⇒dx=2t3−3t2−33t2−6t+3dtThayCận⇒I=b∫a2t3−3t2−33t2−6t+3dt(t2−33t−3−1)tĐếnĐâyrútgọnđilàtốtrồi
a)I=7/9∫2/3dx√(x−1)(x−13)Đặt(x−1)t=√(x−1)(x−13)⇔(x−13)t2=(x−1)⇔x=t2−33t−3⇒dx=2t3−3t2−33t2−6t+3dt$$Thay Cận bạn tự thay\Rightarrow I=\int\limits_{a}^{b}\frac{\frac{2t^3-3t^2-3}{3t^2-6t+3}dt}{(\frac{t^2-3}{3t-3}-1)t}Đến Đây rút gọn đi là tốt rồi$$