How do you integrate #int (2x+1)/((x+8)(x-2)(x-3)) # using partial fractions?

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Answer 1 · 2017-01-31T16:13:55

The answer is #=-3/22ln(|x+8|)-1/2ln(|x-2|)+7/11ln(|x-3|)+C#

Let's perform the decomposition into partial fractions

#(2x+1)/((x+8)(x-2)(x-3))=A/(x+8)+B/(x-2)+C/(x-3)#

#=(A(x-2)(x-3)+B(x+8)(x-3)+C(x+8)(x-2))/((x+8)(x-2)(x-3))#

As the denominators are the same, we can equalise the numerators

#(2x+1)=A(x-2)(x-3)+B(x+8)(x-3)+C(x+8)(x-2)#

Let #x=-8#, #=>#, #-15=110A#, #=>#, #A=-3/22#

Let #x=2#, #=>#, #5=-10B#, #=>#, #B=-1/2#

Let #x=3#, #=>#, #7=11C#, #=>#, #C=7/11#

Therefore,

#(2x+1)/((x+8)(x-2)(x-3))=(-3/22)/(x+8)+(-1/2)/(x-2)+(7/11)/(x-3)#

So,

#int((2x+1)dx)/((x+8)(x-2)(x-3))=-3/22intdx/(x+8)-1/2intdx/(x-2)+7/11intdx/(x-3)#

#=-3/22ln(|x+8|)-1/2ln(|x-2|)+7/11ln(|x-3|)+C#