How do you express #(x+1)/(x^2 + 6x)# in partial fractions?

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How do you express #(x+1)/(x^2 + 6x)# in partial fractions?

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Answer 1 · 2016-03-08T17:20:06

#(x+1)/(x(x+6)) = (1/6)/x + (5/6)/(x+6) = 1/(6x)+5/(6(x+6))#

Explanation:

#(x+1)/(x(x+6)) = A/x + B/(x+6) #
#(x+1)=A(x+6)+Bx#
#x+1 = Ax+6A+Bx#
#1=A+B, 6A=1#
#A=1/6 , B=1-1/6 =5/6#
#(x+1)/(x(x+6)) = (1/6)/x + (5/6)/(x+6) = 1/(6x)+5/(6(x+6))#

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How do you express #(x+1)/(x^2 + 6x)# in partial fractions?

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