To solve this you need to integrate by substitution. You can spot this because the differential of the bottom of the fraction is a multiple of the top part, showing this quickly; if u = x2 + 3 (the bottom part) then du/dx = 2x, which is a multiple of 2 greater than x (the top part). So if we continue using u = x2 + 3 by substituting that into the equation as well as substituting the dx term (at the end of the integral) by using a rearrangement of du/dx = 2x [dx = du/2x]. Thus we are left with: Integral of (x/u).(du/2x), this means we can cancel the x terms out leaving us with (1/2). Integral 1/u.du which will equal (1/2) ln(u), so substituting out u finally gives us (1/2) ln( x2 + 3).