Though, once you become proficient, you can probably skip the substitution and go straight to the integration. This MUST be present or you cannot use this form.įor beginners, using u-substitution to make absolutely sure you have an exact match is the wisest course. This is the derivative the whatever you call u. Of course, we wouldn't typically show all of these steps, i just included them so you could get the idea.īut the big thing to understand is this: In ALL of the standard integral forms there is a du. (15/2) ∫ du/ u ← I now have a standard form and can integrate ∫ 15 dx / (2x+3) ← though not necessary, factor out the 15 to avoid mistakesġ5∫ dx / (2x+3) ← now replace what you have with the u and du.ġ5∫ (½ du) / (2x+3) ← we do this because dx = ½duġ5∫ (½ du) / u ← no we need to make this EXACTLY match the standard form. So, to use this form I would need to do the following: That is fine, but you MUST have its derivative. The du is the derivative of whatever you call u - it MUST be present or you cannot use that standard derivative form.įor the ∫ (1/u) du = ln(u) + C form, you must have the derivative of the denominator you are calling u.įor 15/(2x+3), you declared u to be 2x+3.

This is NOT just some notation that you can ignore, it is vital to getting the correct answer. Notice that in the standard integral forms, you ALWAYS have a du.

Integral khan academy