How do you integrate $int (2x^2+9x-6) / (x^2+x-6)$ using partial fractions?

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Answer 1 · 2016-12-18T18:14:36

The answer is $=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C$

We need to simplify the expression

The denominator is $x^2+x-6=(x+3)(x-2)$

Let's do a long division

$color(white)(aaaa)$ $2x^2+9x-6$ $color(white)(aaaa)$ $∣$ $x^2+x-6$

$color(white)(aaaa)$ $2x^2+2x-12$ $color(white)(aaaa)$ $∣$ $2$

$color(white)(aaaaaa)$ $0+7x+6$

$(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)$

We can now do the decomposition into partial fractions

$(2x+6)/(x^2+x-6)=(7x+6)/((x+3)(x-2))$

$=A/(x+3)+B/(x-2)$

$=(A(x-2)+B(x+3))/((x+3)(x-2))$

Therefore,

$7x+6=A(x-2)+B(x+3)$

Let $x=2$ , $=>$ . $20=5B$ , $=>$ , $B=4$

Let $x=-3$ , $=>$ , $-15=-5A$ , $=>$ , $A=3$

$(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)=2+3/(x+3)+4/(x-2)$

$int((2x^2+9x-6)dx)/(x^2+x-6)=int2dx+3intdx/(x+3)+4intdx/(x-2)$

$=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C$

How do you integrate int (2x^2+9x-6) / (x^2+x-6)int (2x^2+9x-6) / (x^2+x-6) using partial fractions?