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Problem 9 Medium Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int \frac{\sqrt{9x^2 + 4}}{x^2}\ dx $

Answer

$$-\frac{\sqrt{9 x^{2}+4}}{x}+3 \ln (3 x+\sqrt{9 x^{2}+4})+C$$

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Video Transcript

Okay, This question wants us to integrate this function using a table. So if we look in the back of the book, we see that the closest thing to this expression is this formula. So what we need to dio is somehow get this integral to look like this one. So we want a use squared as our function inside the square root. So we want you squared to be equal to nine x squared. So this means that you is equal to three x, so d'you is equal to three d X or D X equals d'you divided by three and some additional things that may come in handy. This means the X squared equals you squared, divided by nine. Okay, so now we can start substituting. So we have are integral of the square root of Well, nine x squared is what we call the you squared plus four, which is two squared, all divided by excess squared, which we said was you squared divided by nine times our chain roll factor of 1/3 d u. And this simplifies to nine divided by three outside the integral, so three times the integral of the square root of you squared plus four over you squared, do you? So now we can directly apply our formula and get three times negative square root you squared plus four divided by you Plus Ln of you plus square root you squared plus a sward But in this case, we already know what a square it iss So let's just write it in to save us a step plus c All right, now let's back Substitute everything so we know that you squared is nine x squared and you is three x so you can see a cancellation is gonna happen here plus the natural log of three x again So we can already see that the three and the 1/3 cancel over there, which gives us a final answer of negative square root nine X squared plus four about my ex, plus the natural log of three x plus the square root of nine x squared plus four plus c. And that's our final answer. Oh, but we need a factor of three here, remember? From right here

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