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

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Answer 1 · 2016-11-19T07:12:51

The answer is #=2/(5x)-7/(5(x+5))#

Let's go to the decomposition in partial fractions

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

#=A/x+B/(x+5)=(A(x+5)+Bx)/(x(x+5))#

so, #(2-x)=A(x+5)+Bx#

Let #x=0#, #=>##2=5A# ; #=># , #A=2/5#

Let #x=-5# ; #=># , #7=-5B# ; #=># , #B=-7/5#

#(2-x)/(x^2+5x)=(2/5)/x+(-7/5)/(x+5)=#