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# 𝘶-substitution: challenging application

Finding  ∫(2^ln x)/x dx. Created by Sal Khan.

## Want to join the conversation?

• There is a much easier way to do this problem, that I came up with.
We know that d/dx (a^x) = a^x lna. Then we can figure out that d/dx (a^x/lna) = a^x because lna is a constant and it can be taken out, then it gets cancelled.
So, antiderivative of a^x dx = a^x/lna !
Then we just have to put u = ln x. We get 2^u which is in the form a^x
• that's also how i would have done this saves a lot of work if you remember that
(1 vote)
• Really dumb question I know, but how come 2^(ln x) = x^(ln 2)
• Not a dumb question.

The key is knowing that 2 = e^ln 2
Review the logarithm properties if you need to on that because that's really useful.

So then
(e^ln 2)^ln x = (e^ln x)^ln 2
exponent properties.

If that part is confusing I like to recall something like
(2^2)^4 = (2^4)^2 = 2^(4*2) = 256

then last we basically repeat the first step
We know e^ln x = x
So
(e^ln x)^ln 2 = x ^(ln 2)

Hope that helps.
• A silly question, but I dont seem to find the answer... is there any difference between u` and du/dx or is it just a simple matter of choice which notation one uses?
• That's not a silly question at all. Generally u' and du/dx are simply alternative notations for the same concept, with u' being the notation that originated with Newton and du/dx originating with Leibniz. You need to know both notations because they're used interchangeably.

But there's more to the story. Newton's notation is more compact and often convenient, such as when we state the product rule, (uv)' = uv' + u'v. Leibniz's is more precise because it specifies that the derivative is with respect to the variable x (which is merely assumed in Newton's notation). It's also more powerful because it invites us to think of du and dx as infinitesimal quantities, which is helpful in understanding integration, and provides a bridge to further topics such as multivariable calculus and differential equations. It's been suggested that mathematicians on the European continent outpaced British mathematicians for many decades following the invention of calculus because they used the more flexible du/dx notation on the continent while the British stuck with u' out of loyalty to Newton.
• Hi, couldn't you use the formula Integral a^x dx = (a^x/ln a) + c
• Yes thats correct. Sal was just explaining the concept without using that formula.
• @ I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?
• e^x and lnx are opposite functions. It would be akin to multiplying by some number and then dividing by that same number. The operations cancel one another out.

e to the natural log of elephant = elephant.
• how to integrate ln(x)^2
• you have to use recursive integration by parts(which is product rule in reverse):
⌠ln(x)^2dx = ln(x)•⌠ln(x)dx - ⌠ln(x)/x dx;
⌠ln(x)dx = x•ln(x) - ⌠x/xdx= x•ln(x) - x (+C) //we plug this back in
⌠ln(x)^2dx = ln(x)•(x•ln(x) - x) - ⌠ln(x)/x dx; u=ln(x), u' = 1/x; dx=du/(1/x);
. . . . . . . . =ln(x)•(x•ln(x) - x) - ⌠1/(x/x)du;
. . . . . . . . =ln(x)•(x•ln(x) - x) - u +C
. . . . . . . . =ln(x)•(x•ln(x) - x)- ln(x) +C
. . . . . . . . =ln(x)•(x•ln(x) - x - 1)+C ; hard, but not as bad as (sec(x))^3.
• You could have taken u=2^ln x differentiated and then substituted right?
• I'm not sure what you're suggesting, but I suspect it wouldn't work. To make a u-substitution work, the formula has to include both the thing you're choosing as u and the derivative of u (that is, du). Sal is able to do a u-substitution using ln x here because the formula also includes 1/x, the derivative of ln x. We can't do a u-substitution using 2^(ln x) because the formula doesn't contain anything corresponding to the derivative of that expression.
• I did this myself, and came up with a different answer...
I converted 2 to e^ln(2) first, so the problem would become
INTEGRAL[e^(ln(2)*ln(x))/x dx] I set u = ln(2)*ln(x) and found du to be ln(2)/x.
So 1/ln(2) * INTEGRAL[e^u dx] = 1/ln(2) * e^u
= e^(ln(2)*ln(x))/ln(2) + C

I'd like to know if this is right as well, and if not -- where I went wrong.
Thank you very much :)
• You have the same answer, you just didn't simplify. Remember that

e^[ log (2) ∙ log (x) ]
= e^[log (2^log(x))]
= 2^log (x)
[2^log x ] / log 2 + C

Please be sure you know your log and exponent properties very thoroughly, as you will use them extensively in calculus.