How do you integrate #int (1-2x^2)/((x-2)(x+6)(x-7)) # using partial fractions?

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Answer 1 · 2016-12-17T19:08:55

The answer is #=7/40ln(∣x-2∣)-71/104ln(∣x+6∣)-97/65ln(∣x-7∣)+C#

Let's do the decomposition into partial fractions

#(1-2x^2)/((x-2)(x+6)(x-7))=A/(x-2)+B/(x+6)+C/(x-7)#

#=(A(x+6)(x-7)+B(x-2)(x-7)+C(x-2)(x+6))/((x-2)(x+6)(x-7))#

Therefore,

#1-2x^2=A(x+6)(x-7)+B(x-2)(x-7)+C(x-2)(x+6)#

Let #x=2#, #=>#, #-7=-40A#, #=>#, #A=7/40#

Let #x=-6#, #=>#, #-71=104B#, #=>#. #B=-71/104#

Let #x=7#, #=>#, #-97=65C#, #=>#, #C=-97/65#

So,

#(1-2x^2)/((x-2)(x+6)(x-7))=(7/40)/(x-2)-(71/104)/(x+6)-(97/65)/(x-7)#

#int((1-2x^2)dx)/((x-2)(x+6)(x-7))=int(7/40dx)/(x-2)-int(71/104dx)/(x+6)-int(97/65dx)/(x-7)#

#=7/40ln(∣x-2∣)-71/104ln(∣x+6∣)-97/65ln(∣x-7∣)+C#