Let's perform the decomposition into partial fractions
(x+1)/((x+5)(x+3)(x+4))=A/(x+5)+B/(x+3)+C/(x+4)x+1(x+5)(x+3)(x+4)=Ax+5+Bx+3+Cx+4
=(A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3))/((x+5)(x+3)(x+4))=A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3)(x+5)(x+3)(x+4)
The denominators are the same, we compare the numerators
(x+1)=(A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3))(x+1)=(A(x+3)(x+4)+B(x+5)(x+4)+C(x+5)(x+3))
Let x=-5x=−5, =>⇒,
-4=-2*-1*A−4=−2⋅−1⋅A, =>⇒, A=-2A=−2
Let x=-3x=−3, =>⇒,
-2=2*1*B−2=2⋅1⋅B, =>⇒, B=-1B=−1
Let x=-4x=−4, =>⇒,
-3=1*-1*C−3=1⋅−1⋅C, =>⇒, C=3C=3
So,
(x+1)/((x+5)(x+3)(x+4))=-2/(x+5)-1/(x+3)+3/(x+4)x+1(x+5)(x+3)(x+4)=−2x+5−1x+3+3x+4
int((x+1)dx)/((x+5)(x+3)(x+4))=-2intdx/(x+5)-1intdx/(x+3)+3intdx/(x+4)∫(x+1)dx(x+5)(x+3)(x+4)=−2∫dxx+5−1∫dxx+3+3∫dxx+4
=-2ln|(x+5)|-ln|(x+3)|+3ln|(x+4)|+C=−2ln|(x+5)|−ln|(x+3)|+3ln|(x+4)|+C
