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

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Answer 1 · 2016-05-13T10:30:19

#int (x+2) /(2x^2-x-1) * d x=-1/2 l n(2x+1)+ l n(x-1)+ C#

#int (x+2) /(2x^2-x-1) * d x=?#

#int (x+2)/((2x+1)*(x-1))d x#

#(x+2)/((2x+1)*(x-1))=A/(2x+1)+B/(x-1)#

#(x+2)/((2x+1)*(x-1))=(A(x-1)+B(2x+1))/((2x+1)(x-1))#

#x=1 rarr" "1+2=A(1-1)+B(2*1+1)#

#3=0+3B" "B=1#

#x=-1/2 rarr" "-1/2+2=A(-1/2-1)+0#

#3/2=A(-3/2)" "A=-1#

#int (x+2) /(2x^2-x-1) * d x=-int (d x)/(2x+1)+int (d x)/(x-1)#

#u=2x+1" "d u=2d x ,d x=(d u)/2" ; "v=x-1" "d v=d x#

#int (x+2) /(2x^2-x-1) * d x=-1/2 int (d u)/u +int (d v)/v#

#int (x+2) /(2x^2-x-1) * d x=-1/2 l n u +l n v +C#

#int (x+2) /(2x^2-x-1) * d x=-1/2 l n(2x+1)+ l n(x-1)+ C#