Let's do the decomposition into partial fractions
(1-2x^2)/((x-2)(x+6)(x-7))=A/(x-2)+B/(x+6)+C/(x-7)1−2x2(x−2)(x+6)(x−7)=Ax−2+Bx+6+Cx−7
=(A(x+6)(x-7)+B(x-2)(x-7)+C(x-2)(x+6))/((x-2)(x+6)(x-7))=A(x+6)(x−7)+B(x−2)(x−7)+C(x−2)(x+6)(x−2)(x+6)(x−7)
Therefore,
1-2x^2=A(x+6)(x-7)+B(x-2)(x-7)+C(x-2)(x+6)1−2x2=A(x+6)(x−7)+B(x−2)(x−7)+C(x−2)(x+6)
Let x=2x=2, =>⇒, -7=-40A−7=−40A, =>⇒, A=7/40A=740
Let x=-6x=−6, =>⇒, -71=104B−71=104B, =>⇒. B=-71/104B=−71104
Let x=7x=7, =>⇒, -97=65C−97=65C, =>⇒, C=-97/65C=−9765
So,
(1-2x^2)/((x-2)(x+6)(x-7))=(7/40)/(x-2)-(71/104)/(x+6)-(97/65)/(x-7)1−2x2(x−2)(x+6)(x−7)=740x−2−71104x+6−9765x−7
int((1-2x^2)dx)/((x-2)(x+6)(x-7))=int(7/40dx)/(x-2)-int(71/104dx)/(x+6)-int(97/65dx)/(x-7)∫(1−2x2)dx(x−2)(x+6)(x−7)=∫740dxx−2−∫71104dxx+6−∫9765dxx−7
=7/40ln(∣x-2∣)-71/104ln(∣x+6∣)-97/65ln(∣x-7∣)+C=740ln(∣x−2∣)−71104ln(∣x+6∣)−9765ln(∣x−7∣)+C
