Let's do the decomposition into partial fractions
(2x+1)/((x-4)(x-1)(x+7))=A/(x-4)+B/(x-1)+C/(x+7)2x+1(x−4)(x−1)(x+7)=Ax−4+Bx−1+Cx+7
=(A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4))/((x-4)(x-1)(x+7))=A(x−1)(x+7)+B(x−4)(x+7)+C(x−1)(x−4)(x−4)(x−1)(x+7)
Therefore,
2x+1=A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4)2x+1=A(x−1)(x+7)+B(x−4)(x+7)+C(x−1)(x−4)
Let x=1x=1, =>⇒,3=-24B3=−24B, =>⇒, B=-1/8B=−18
Let x=4x=4, =>⇒,9=33A9=33A, =>⇒, A=3/11A=311
Let x=-7x=−7, =>⇒, -13=88C−13=88C, =>⇒C=-13/88C=−1388
So,
(2x+1)/((x-4)(x-1)(x+7))=(3/11)/(x-4)-(1/8)/(x-1)-(13/88)/(x+7)2x+1(x−4)(x−1)(x+7)=311x−4−18x−1−1388x+7
int((2x+1)dx)/((x-4)(x-1)(x+7))=3/11intdx/(x-4)-1/8intdx/(x-1)-13/88intdx/(x+7)∫(2x+1)dx(x−4)(x−1)(x+7)=311∫dxx−4−18∫dxx−1−1388∫dxx+7
=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C=311ln(∣x−4∣)−18ln(∣x−1∣)−1388ln(∣x+7∣)+C
