 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Substitutions Involving e^x or ln(x) - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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To take the integral of functions involving e raised to some power, substitute the power that e is raised to with w. Then, use the substitution method: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.

We're asked to perform the antidifferentiation on this integral. The integral of e to the -x over 1 plus e to the -x. Believe or not, this integral is going to involve, if you follow me over here, this integral formula. So by a change of variables, I can get this kind of an integral out of the one that I started with. So let's see how that works.

Now sometimes when you're using the method of substitution. You have to think in terms of functions and their derivatives. You really need to be good at your derivative formulas in order to use the method of substitution well. Here I'm seeing that this denominator, 1 plus e to the -x, its derivative, remember if you differentiate it with this whole denominator, the derivative of the 1 would be 0. You'd get the derivative of this being -e to the -x which is very similar to what I have on top.

So you're often looking for a function and its derivative in your intergrand, when you're using the method of substitution. So I'm going to substitute for the denominator; 1 plus e to the -x. Then, when I differentiate this, the derivative of e to the -x is -e to the -x. Then I have a dx. This is not exactly what I have in the numerator. I have a positive e to the -x, but I can just multiply both sides by -1. That gives me e to the -x dx.

So let me make the substitution. The top becomes -dw, and the bottom becomes just w. So this is the same as negative integral 1 over w, dw. This is exactly what I was hoping for. An integral whose formula I know. Now the integral of 1 over w, dw is natural log of the absolute value of w plus c. So I have the negative in the front, but natural log of the absolute value w. All that's left to be done is you replace the w, with 1 plus e to the -x. So this is natural log of 1 plus e to the -x plus c. Perfect. Done.

So I'm not going to check this by differentiation, but you could. If you want to check your answer, that's the great thing about antidifferentiating, you can just differentiate to check it. You should get this; that you started with this function.

Let's go to another example very similar. Similar in the sense that, our substitution will yield a similar integral, but doesn't look similar at the beginning. Now here again, the advice; you've got to look for a function and its derivative together in the intergrand, what's inside the integral. Here I'm seeing that natural log has the derivative 1 over x. I have a 1 over x. This is my 1 over x. So those two kind of go together. I want to substitute for the function whose derivative I also see. So w equals natural log x. Dw equals 1 over x dx. That's exactly what I have here, dx over x, that's my dw. So I have dw over, and this guy is my w. That's my dw, and this is my w.

So this is exactly the integral of 1 over w, dw. That's going to give me natural log of the absolute value of w plus c. The very unexpected answer when you substitute for w, you get natural log x inside. Natural log of the absolute value of natural log x plus c. Perfectly good function. Graph it. That's the answer to this problem; the integral of dx over x lnx. Natural log of the absolute value of natural log x plus c.