 ###### 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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# Simple Substitutions - 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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When taking the integral of something in the form 1/(some function of x), apply the rule that the antiderivative of 1/x is ln|x|. The expression of x will be what is substituted with w. Then, use the method of substitution: 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 take a look at another example. I want to solve this indefinite integral. The integral of 6 over 2x plus 1. This looks a little bit like something of the form 1 over x dx, but you want to be careful about just applying these antiderivative formula; ln of the absolute value of x plus c, to a problem like this.

Anytime that you look at an integral and you see a composite function, and here I think I have 6 over 2x plus 1 is a composite function with the inside function, 2x plus 1. Anytime you have a composite function, you really should use the method of substitution. Just as if you were differentiating this, you’d use the chain rule.

Let’s substitute for the inside part which is 2x plus 1. And then dw is 2 times dx. Now I have a 6 times dx and so I can just write this as 3dw over 2x plus 1, that’s w. Let me pull the 3 outside. I have 3 times 1 over wdw, and of course the integral of 1 over w is natural log. So it’s 3 times natural log of the absolute value of w. And then I re-substitute. W is 2x plus 1, 3 natural log absolute value of 2x plus 1 plus c. And that’s my answer. Now let me just check this by differentiating it and if I get this, then I’m all set, this will be the correct answer.

Derivative with respect to x of 3 times the natural log of absolute value of 2x plus 1, plus c. I could pull the 3 out in front; the derivative of the natural log of the absolute value of something is just one over that something. But I have to multiply by the derivative of that something. So the derivative of this is 2. And the derivative of the plus c is 0. This becomes 6 over 2x plus 1, and that’s exactly what I started with. So that means that my answer, 3 natural log of the absolute value of 2x plus 1 plus c is correct.