How do you integrate #int (1-x^2)/((x+1)(x-5)(x+2)) # using partial fractions?

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Answer 1 · 2016-12-27T08:08:14

The answer is #=-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C#

#1-x^2=(1+x)(1-x)#

Let's rewrite the expression

#(1-x^2)/((x+1)(x-5)(x+2))=(cancel(1+x)(1-x))/(cancel(x+1)(x-5)(x+2))#

Now we can do the decomposition into partial fractions

#(1-x)/((x-5)(x+2))=A/(x-5)+B/(x+2)#

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

Therefore,

#1-x=A(x+2)+B(x-5)#

Let #x=-2#, #=>#, #3=-7B#, #=>#, #B=-3/7#

Let #x=5#, #=>#, #-4=7A#, #=>#, #A=-4/7#

So,

#(1-x)/((x-5)(x+2))=(-4/7)/(x-5)+(-3/7)/(x+2)#

Therefore,

#int((1-x)dx)/((x-5)(x+2))=int((-4/7)dx)/(x-5)+int((-3/7)dx)/(x+2#

#=-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C#