evaluate the integral from 4 to 3.
x/2x^2-6dx
I will assume that you meant:
∫ x/(2x^2 - 6) dx from 3 to 4 , (not 4 to 3)
= (1/4) ln(2x^2 - 6) from 3 to 4
= (1/4) (ln 26 - ln 12)
= (1/4) ln (26/12)
= (1/4) ln(13/6) or (1/4)(ln 13 - ln6)
x/2x^2-6dx
∫ x/(2x^2 - 6) dx from 3 to 4 , (not 4 to 3)
= (1/4) ln(2x^2 - 6) from 3 to 4
= (1/4) (ln 26 - ln 12)
= (1/4) ln (26/12)
= (1/4) ln(13/6) or (1/4)(ln 13 - ln6)