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# Substitutions Involving e^x or ln(x) - Problem 2

###### 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.

When looking at a function involving 1 over something, use substitution for replace the "something" being divided. Recall that the antiderivative of 1/x is ln|x|. 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.

Let's do another method of substitution problem. In this problem, we're going to solve two integrals, and both of them are going to be in the form, rational functions. For both of them I'm going to use this formula if you follow me over here; The integral of 1 over x dx equals natural log of the absolute value of x plus c.

Now first of all, let me caution you not to use that formula directly. You don't want to just write equals natural log of the absolute value of 2 minus 5x plus c. That's not going to work. In this case, you want to use substitution. You want to remember to use substitution anytime if you were differentiating. Use the chain rule. This is a composite function, it would need the chain rule for differentiation. So you'd use substitution to integrate it. So let's see how that works.

I think of this as like a 1 over x function with 2 minus 5x inside it. So the inside function is 2 minus 5x. That's what I'm going to substitute for. I need a dw or I need to replace the dx with something. So dw is going to be -5dx.

So if I solve this for dx, I get -1/5dw. So let me replace the dx with -150w. This integral becomes -1/5dw over, and 2 minus 5x is just w. Now if I pull the -1/5 out, look what I have. I have the integral of dw over w which is really just the same as the integral of 1 over w dw. So this is exactly the integral I have in my formula. I can apply that formula now. -1/5 natural log of the absolute value of w plus c. The only thing left to do is replace the w with 2 minus 5x. So this is going to be -1/5 natural log of the absolute value of 2 minus 5x plus c. That's my final answer.

So let's check this by differentiating it. If we differentiate it, we should get the original function; 1 over 2 minus 5x back again. So let's see if that happens. So the derivative of -1/5 natural log of the absolute value of 2 minus 5x plus c is, I can pull out the -1/5.

The derivative of the natural log of the absolute value of something is just 1 over that something. So this is going to be 1 over 2 minus 5x. This is a composite function so I have to multiply by the derivative of the inside part. The derivative of the inside part is -5, so times -5. You can see that these are going to cancel, the -1/5 and the -5 cancel. I get 1 over 2 minus 5x. That's exactly what I started with. That tells me that this answer is correct

Let's try another one. This one is slightly harder. It's harder to see that we're going to get something of the form 1 over w dw here, but keep in mind that this does involve a composite function of the form 1 over where w would be 9 plus x². So let's actually try substituting for that 9 plus x². So dw would be 2xdx.

Now I have an xdx. I have six of them, so that's going to be 3 times dw. So let me substitute for that. 6xdx is 3 times this so 3dw. Then the 9 plus x², that's just w. So I'm in the same situation as before. This integral can be written if I pull the 3 out. The integral of 1 over w dw. On this I can apply my formula.

So this becomes 3 times the natural log of the absolute value of w plus c. The final thing I have to do, replace the w with 9 plus x². 3 times the natural log of the absolute value of 9 plus x² plus c. Now you might notice that in this final answer, I don't absolutely need the absolute value, because 9 plus x² is never going to be negative. It's actually a quantity that can't be negative, because x² can't be negative. This is always going to be greater than or equal to 9. I'll leave it. It's redundant, but it's not incorrect.

Let's differentiate this to see if it's the correct answer. If it's the correct answer differentiating it should give me 6x over 9 plus x². So let's do that. Now I can pull the 3 out, and I have the derivative of the natural log of the absolute value of something. That's just 1 over that something. So 1 over the 9 plus x² times the derivative of the inside part. I have to use the chain rule on this. The derivative of the inside part is 2x. That's exactly what I was hoping for; 3 times 2x is 6x. 6x over 9 plus x². That's exactly what I started with. That means that this answer is correct.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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