# What is the definision of lan transformans

1. Feb 17, 2007

### transgalactic

i can take every time all the function on the buttom
put it in a lan
an devide it in its derivative?

even if i got a fuction
1/(x+2)^2

its integral solves [(x+2)^-1]/-1

or i can open this equetion (a+b)^2=x^2+2*x*2+2^2
then put it all in a lan ln(x^2+2*x*2+2^2) and i will devide it by it derivative

2x+4

ln(x^2+2*x*2+2^2)/2x+4

what is my red sigh
how do i deside to what way to go???

2. Feb 17, 2007

### HallsofIvy

Staff Emeritus
I hate to criticize anyone's English (since my (put the language of your choice here!) is awful) but I can't really understand what you are saying here. In particular, what do you mean by " lan"?

Here is my response to what I think you are asking:
Yes, it is true that the anti-derivative, or integral, of (x+2)-2 is
(x+2)-1 divided by -1: that's the power rule: -2+ 1= -1 (and it helps that the derivative of x+ 2 itself is 1).

If you had (3x+ 2)-2 to integrate, you can do much the same thing: (3x+2)-1 but now divided by the derivative of 3x+ 2 which is 3: (-1/3)(3x+2)-1. That's because if you differentiate (-1/3)(3x+2)-1, you would use both the "power rule" (with n-1= -1-1= -2) and the "chain rule" (the derivative of 3x+2 is 3) to multiply by -1 and 3 and change the power to -2: (-1)(3)(-1/3)(3x+2)-2, the original function.

However it is important there that the "inner" function, x+ 2 in the first case and 3x+2 in the second, is linear and its derivative is a constant.

The "chain rule in reverse" is really "integration by substitution": To integrate (3x+2)-2, more formally, let u= 3x+ 2. Then du= 3 dx or dx= (1/3)du and (3x+2)-2= u-2 so $\int (3x+2)^{-2}dx= (1/3)\int u^{-2}du= (-1/3)u^{-1}+ C= (-1/3)(3x+2)^{-1}+C$.

If that were (x2+ 2)-2 we cannot do that! If we try to let u= x2+ 2 then du= 2x dx and there is NO "2x" in the integral. While we could write (1/3)du= dx before, we cannot write (1/2x)du= dx because we can't have an "x" in the "u" integral.

In particular, to integrate (x2+ 4x+ 4)-1, which I think is your second example, we CANNOT just say "well since that is a -1 power, the anti-derivative is a logarithm and the we divide by the derivative of x2+ 4x+ 4= 2x" because that derivative is NOT a constant.
If you tried to make the substitution u= x2+ 4x+ 4, then du= (2x+ 4) dx and we cannot just divide by that!

Of course, you can always check an integral by differentiating;
What do you get if you differentiate ln(x2+ 4x+ 4)/(2x+ 4)?

You would have to use the quotient rule: the derivative of ln(x2+ 4x+ 4) time 2x+4 minus ln(x2+ 4x+ 4) times the derivative of 2x+ 4, all divided by (2x+4)2. That is
{(1/(x^2+ 4x+ 4)(2x+ 4)(2x+4)- ln(x2+ 4x+ 4)(2)}/(2x+4)2. That is not anything like your original (x2+ 4x+ 4)-1!!

The rule "integrate f(x)n by integrating un and then dividing by the derivative of f works only if f is linear so its derivative is a constant. Otherwise see if its derivative is already in the integral so you can substitute.

Last edited: Feb 17, 2007
3. Feb 17, 2007

### transgalactic

so i wich case we know that an integral equals to a ln fuction

and what are my options in the case of non lenear fraction like x^2+1
???

i know that one way is using trigonometric substitution

but what do i do i a case of e^x for example????

Last edited: Feb 17, 2007