Anon481516 said:
Alright. But I noticed that there are many rules regarding integrating a function. Some rules look similar to one another, but you can only use one rule at a time, correct? From the previous example, you went from the actual function to 1/2 ∫ (2x-5) ... (2dx), then from there you used the n+1/n+1 rule. Some of these rules can seem and are somewhat basic, but how do you know what rules to use? Do People have a sheet of rules by them when doing these? Once functions use logarithms or square roots, they will involve natural logs and other such equivalents when integrating, then they will get rather complex and I don't see how people could do them from memory alone.
In my opinion, the most important skill you need for doing integrations is to be very accomplished at taking derivatives, in particular, being able to use the chain rule several levels deep and also being good with the product rule, as well as with the sum and difference rules. You should
know the derivatives of all the basic functions. When I say
know I mean you should be able to write them down without looking them up.
If you are familiar with integral tables, (Does anybody use these any more?) you will notice that rather than being in the form
\displaystyle \int\,x^n\,dx=<br />
\frac{1}{n+1}x^{n+1}+C
and
\displaystyle \int\,\cos(x)\,dx=<br />
\sin(x)+C
they are in the form
\displaystyle \int\,u^n\,du=\frac{1}{n+1}u^{n+1}+C
and
\displaystyle \int\,\cos(u)\,du=<br />
\sin(u)+C\ .
For the integral in your example, you have (2x-5)
3 as the integrand. In this case, u = (2x-5), therefore, du = u'(x)dx = 2dx . (That can also be written, du = d(2x - 5) = 2dx .
So they multiplied dx by (1/2)∙2, kept the 2 with the dx and move the constant multiplier out of the integral. Then, (2x-5)
3 (2dx) becomes (2x-5)
3d(2x-5) which is u
3du. Now you can use the power rule for the anti-derivative (the integral).
