Please help me evaluate, ∫ ( (15x^2 + 6x - 9) / (3x) )dx

I always give 10 points to correct answers

integral (-9+6 x+15 x^2)/(3 x) dx

Factor out constants

= 1/3 integral (15 x^2+6 x-9)/x dx

For the integrand (15 x^2+6 x-9)/x, do long division:
= 1/3 integral (15 x-9/x+6) dx


= 1/3 integral 6 dx-3 integral 1/x dx+5 integral x dx


= 2 x-3 integral 1/x dx+5 integral x dx

The integral of 1/x is log(x):

= 2 x-3 log(x)+5 integral x dx

= (5 x^2)/2+2 x-3 log(x)+constant

∫ (15x² + 6x − 9) / (3x) dx

= ∫ (15x²/(3x) + 6x/(3x) − 9/(3x)) dx

= ∫ (5x + 2 − 3/x) dx

= 5x²/2 + 2x − 3ln(x) + C

http://jsfiddle.net/Sxv7f/1/embedded/res…