How do you integrate $int (2x+1)/(4x^2+12x-7) dx$ using partial fractions?

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Answer 1 · 2016-10-26T15:04:45

The integral $int((2x+1)dx)/(4x^2+12x-7)=3/8ln(2x+7)+1/8ln(2x-1)+C$

Let's factorise the denominator
$4x^2+12x-7=(2x+7)(2x-1)$

We look for the partial fractions
$(2x+1)/(4x^2+12x-7)=A/(2x+7)+B/(2x-1)$

$(2x+1)/(4x^2+12x-7)=(A(2x-1)+B(2x+7))/(4x^2+12x-7)$

so, $2x+1=A(2x-1)+B(2x+7))$
Let $x=1/2$ $=>$ $2=8B$ $=>$ $B=1/4$
Let $x=0$ $=>$ $1=-A+7/4$ $=>$ $A=3/4$

$(2x+1)/(4x^2+12x-7)=(3/4)/(2x+7)+(1/4)/(2x-1)$
Let's do the integral
$int((2x+1)dx)/(4x^2+12x-7)=int((3/4)dx)/(2x+7)+int((1/4)dx)/(2x-1)$

$=3/8ln(2x+7)+1/8ln(2x-1)+C$

How do you integrate int (2x+1)/(4x^2+12x-7) dxint (2x+1)/(4x^2+12x-7) dx using partial fractions?