lets say the function was (2x^2+6x)^3 .

and to integrate this , u would get..


(2x^2+6x)^4
-----------
4(4x+6)

right?

u just add one to the power, and divide by the power times the derivative of whats inside the paranthesis?

what is your u in the u substitution? and more importantly, what is your du?

easy way to test it is to take the derivative of your answer and you should get the original expression back.

This post was edited by carteblanche on Nov 9 2014 10:52pm

It's not so simple. You want to integrate f(x)^3, it is not equal to f(x)^4/f '(x).
However you can write f(x) = x^3 * (2x+6)^3 and integrate by part 3 times. In this case g(x) = (2x+6)^3 is easy to integrate: int g(x) = 1/8*(2x+6)^4, and so on.

Would be far easier to just develop (2x^2+6x)^3 and then integrate term by term...

So when is the case where I can do f(x)^4/f '(x)? And how do u integrate this witout simplifying this expression?

you're trying to use u substitution. that means for your u, you need du in there.

f(x) = (2x^2+6x)^3

u = 2x^2 + 6x
du = 4x + 6

g(u) = u^3

notice how you're not using du? you need g(u) = u^3 du

so if the original function was:
f(x) = (2x^2+6x)^3 * (4x+6)

it would be fine.

or if du = 1, thats the same thing.
eg:
f(x) = (x + 5)^5
u = (x + 5)
du = 1
g(u) = u^5 du

so you can do what you wanted.

so this "f(x)^4/f '(x)" is basically u-du substituion?

so theres no other method of solving this kind of integration without expanding that expression ?

If you can't find a semi-okay way to do integration by parts, there isn't another way to integrate this by hand other than expansion.

Polynomial functions is easy to integrate term by term.