How do you integrate #int(x+1)/((x+5)(x+3)(x+4))# using partial fractions?

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Answer 1 · 2017-06-16T07:31:41

The answer is #=-2ln|(x+5)|-ln|(x+3)|+3ln|(x+4)|+C#

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)#

#=(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))#

Let #x=-5#, #=>#,

#-4=-2*-1*A#, #=>#, #A=-2#

Let #x=-3#, #=>#,

#-2=2*1*B#, #=>#, #B=-1#

Let #x=-4#, #=>#,

#-3=1*-1*C#, #=>#, #C=3#

So,

#(x+1)/((x+5)(x+3)(x+4))=-2/(x+5)-1/(x+3)+3/(x+4)#

#int((x+1)dx)/((x+5)(x+3)(x+4))=-2intdx/(x+5)-1intdx/(x+3)+3intdx/(x+4)#

#=-2ln|(x+5)|-ln|(x+3)|+3ln|(x+4)|+C#