How do you integrate #int (x+5)/((x+3)(x-2)(x-7)) # using partial fractions?

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Answer 1 · 2017-10-05T16:42:33

The answer is #=1/25ln(|x+3|)-7/25ln(|x-2|)+6/25ln(|x-7|)+C#

Perform the decomposition into partial fractions

#(x+5)/((x+3)(x-2)(x-7))=A/(x+3)+B/(x-2)+C/(x-7)#

#=(A(x-2)(x-7)+B(x+3)(x-7)+C(x+3)(x-2))/((x+3)(x-2)(x-7))#

The denominators are the same, compare the numerators

#(x+5)=(A(x-2)(x-7)+B(x+3)(x-7)+C(x+3)(x-2))#

Let #x=-3#, #=>#, #2=50A#, #A=1/25#

Let #x=2#, #=>#, #7=-25B#, #B=-7/25#

Let #x=7#, #=>#, #12=50C#, #C=6/25#

Therefore,

#(x+5)/((x+3)(x-2)(x-7))=(1/25)/(x+3)+(-7/25)/(x-2)+(6/25)/(x-7)#

So, the integral is

#int((x+5)dx)/((x+3)(x-2)(x-7))=int(1/25dx)/(x+3)+int(-7/25dx)/(x-2)+int(6/25dx)/(x-7)#

#=1/25ln(|x+3|)-7/25ln(|x-2|)+6/25ln(|x-7|)+C#