Calculas: Antiderrivatives?

Question

need to find the antiderivative of: ln|x|

I think that it would be just (x) or even (1/2x^1/2),
since ive added to eliminate
natural logs in the past, when both sides where natural logs.... i doubt it though since this is only an expression.

Answer 1

The antiderivative of ln(x) is a function f(x) such that, when you differentiate it, you get ln(x).

Well, if you have trouble, work toward an answer stepwise.

The product rule is our friend; the derivative of xln(x) will be close to what we want, although not exactly right.

Let g(x) = xln(x). The g'(x) = ln(x) + 1.

So if f is the right answer, (f-g)' = -1. i.e., f-g = -x. f(x) = xln(x) - x.

There's your answer, although you'd have to rewrite it to make it suitable for turning in.

Answer 2

You were asking for the anti-derivative of ln |x|.

The first answer is just wrong because a very necessary term in x was omitted --- the constant C just doesn't cut it; and the second answer isn't so obviously complete because it didn't deal with the issue of the argument being mod (x) or |x|. THAT'S rather important!


One danger in dealing with the integration or differentiation of functions of |x| is that it may not be obvious when POWERS of x that come in to a solution should really be plain ' x ' or ' |x| ' !

You can try to resolve such difficulties for yourself by building up a mental library of how to "do d/dx" of some representative functions or powers of |x|.

Another way is to simply grit one's teeth and work through the two separate cases, where (i) x > 0, and (ii) x < 0. It's worth understanding the general method for that. Here's what you do, with ln |x| as an example.

For example, when x < 0 :

d/dx of ln |x| = d/dx [ ln (-x)] = - d/dz [ln z] where z ( = - x) > 0.

Now you're on safe ground, with the ln argument, z, > 0. You can now work out what the derivative is, converting it back to what it looks like in terms of x and/or |x| separately at the end.

In your case, when x < 0, it's worth starting with checking out
d/dx [x ln |x|], converting that to d/dx [x ln (-x)], and seeing what results with the substitution z = -x so that the ' ln ' term then has a positive argument:

d/dx [x ln |x|] = d/dx [x ln (-x)] = - d/dz [- z ln z] {for z > 0}

= + d/dz [z ln z] = 1 + ln z = 1 + ln |x|, so:

d/dx [x (ln |x| - 1)] = (1 + ln |x|) - 1 = ln |x|.

I'm presuming that the corresponding argument for x > 0 gives you no difficulty --- just repeat the above without having to worry about "moding" the ' x ,' or see Curt Monash's argument (done only for positive x, I'm afraid) just above, for that.

That means that we're finally home and dry! :

The antiderivative of ln |x| = x (ln |x| - 1) .

Note here what I warned you about: only the ' x' that is in the argument as |x| is in fact "modded." The ' x ' OUTSIDE the parenthesis ISN'T "modded."

I hope that by explaining how to resolve any possible such ambiguities with "mod" functions or arguments, I have equipped you to deal with them yourself in future.

Good luck !

Live long and prosper.

Answer 3

If you integrate by parts taking the integral of dx firstyou´ll get

x*lnIxI-Int dx = x*(lnIxI-1)

Answer 4

The integral of in X = X in X + C

http://en.wikipedia.org/wiki/Table_of_integrals

This explains them all