∫x2−8x+4x3−4xdxLet x2−8x+4x3−4x=x2−8x+4x(x2−4)
=x2−8x+4x(x−2)(x+2)
Let x2−8x+4x(x−2)(x+2)=Ax+Bx−2+Cx+2
⇒x2−8x+4=A(x2−4)+B(x2+2x)+C(x2−2x)⇒x2−8x+4=(A+B+C)x2+(2B−2c)x−4A
On comparing,
A+B+C=1,2B−2C=−8,−4A=4
On solving, A=−1,B=−1,C=3
∴∫x2+8x+4x3−4xdx=−1∫1xdx−∫1(x−2)dx
=−log|x|−log|x−2|+3log|x+2|+K