How do you express #(x+1)/(x^2 + 6x)# in partial fractions? Precalculus Matrix Row Operations Partial Fraction Decomposition (Linear Denominators) 1 Answer Bdub Mar 8, 2016 #(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))# Answer link Related questions What does partial-fraction decomposition mean? What is the partial-fraction decomposition of #(5x+7)/(x^2+4x-5)#? What is the partial-fraction decomposition of #(x+11)/((x+3)(x-5))#? What is the partial-fraction decomposition of #(x^2+2x+7)/(x(x-1)^2)#? How do you write #2/(x^3-x^2) # as a partial fraction decomposition? How do you write #x^4/(x-1)^3# as a partial fraction decomposition? How do you write #(3x)/((x + 2)(x - 1))# as a partial fraction decomposition? How do you write the partial fraction decomposition of the rational expression #x^2/ (x^2+x+4)#? How do you write the partial fraction decomposition of the rational expression # (3x^2 + 12x -... How do you write the partial fraction decomposition of the rational expression # 1/((x+6)(x^2+3))#? See all questions in Partial Fraction Decomposition (Linear Denominators) Impact of this question 1290 views around the world You can reuse this answer Creative Commons License