We need to simplify the expression
The denominator is x^2+x-6=(x+3)(x-2)x2+x−6=(x+3)(x−2)
Let's do a long division
color(white)(aaaa)aaaa2x^2+9x-62x2+9x−6color(white)(aaaa)aaaa∣∣x^2+x-6x2+x−6
color(white)(aaaa)aaaa2x^2+2x-122x2+2x−12color(white)(aaaa)aaaa∣∣22
color(white)(aaaaaa)aaaaaa0+7x+60+7x+6
(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)2x2+9x−6x2+x−6=2+7x+6x2+x−6
We can now do the decomposition into partial fractions
(2x+6)/(x^2+x-6)=(7x+6)/((x+3)(x-2))2x+6x2+x−6=7x+6(x+3)(x−2)
=A/(x+3)+B/(x-2)=Ax+3+Bx−2
=(A(x-2)+B(x+3))/((x+3)(x-2))=A(x−2)+B(x+3)(x+3)(x−2)
Therefore,
7x+6=A(x-2)+B(x+3)7x+6=A(x−2)+B(x+3)
Let x=2x=2, =>⇒. 20=5B20=5B, =>⇒, B=4B=4
Let x=-3x=−3, =>⇒, -15=-5A−15=−5A, =>⇒, A=3A=3
(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)=2+3/(x+3)+4/(x-2)2x2+9x−6x2+x−6=2+7x+6x2+x−6=2+3x+3+4x−2
int((2x^2+9x-6)dx)/(x^2+x-6)=int2dx+3intdx/(x+3)+4intdx/(x-2)∫(2x2+9x−6)dxx2+x−6=∫2dx+3∫dxx+3+4∫dxx−2
=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C=2x+3ln(∣x+3∣)+4ln(∣x−2∣)+C
