∫6x2−17x−5(x−3)(x−2)2dx
Let 6x2−17x−5(x−3)(x−2)2=Ax−3+B(x−2)+C(x−2)2
⇒6x2−17x−5=A(x−2)2+B(x−3)(x−2)+C(x−3)
=A(x2−4x+4)+B(x2−5x+6)+C(x−3)
Comparing Coefficients,
A+B=6
−4A−5B+C=−17
4A+6B−3C=−5
⇒A=−2,B=8,C=15
⇒∫6x2−17x−5(x−3)(x−2)2dx=∫(−2x−3+8x−2+15(x−2)2)dx
=−2log|x−3|+8log|x−2|−15(x−2)+C