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

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

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

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

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

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

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

$3=0+3B" "B=1$

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

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

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

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

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

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

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

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