How do you integrate #int (2x+1)/((x-4)(x-1)(x+7)) # using partial fractions?

Question

1645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T10:44:38

The answer is #=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C#

Let's do the decomposition into partial fractions

#(2x+1)/((x-4)(x-1)(x+7))=A/(x-4)+B/(x-1)+C/(x+7)#

#=(A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4))/((x-4)(x-1)(x+7))#

Therefore,

#2x+1=A(x-1)(x+7)+B(x-4)(x+7)+C(x-1)(x-4)#

Let #x=1#, #=>#,#3=-24B#, #=>#, #B=-1/8#

Let #x=4#, #=>#,#9=33A#, #=>#, #A=3/11#

Let #x=-7#, #=>#, #-13=88C#, #=>##C=-13/88#

So,

#(2x+1)/((x-4)(x-1)(x+7))=(3/11)/(x-4)-(1/8)/(x-1)-(13/88)/(x+7)#

#int((2x+1)dx)/((x-4)(x-1)(x+7))=3/11intdx/(x-4)-1/8intdx/(x-1)-13/88intdx/(x+7)#

#=3/11ln(∣x-4∣)-1/8ln(∣x-1∣)-13/88ln(∣x+7∣)+C#