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Numerade Educator



Problem 49 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{1}{x \sqrt{4x + 1}}\ dx $


$\ln \left|\frac{\sqrt{4 x+1}-1}{\sqrt{4 x+1}+1}\right|+C$


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Video Transcript

Let's start off with the use of beer. Let's just take you to be that radical in the denominator square root for it's plus one. Go ahead and usar calculus or differentiation skills to differentiate this and simple fire. It's for X plus one inside the radical in the bottom, two up top and I could rewrite This is too over you DX and then taking this equation here you get the eggs by itself, so that gives you you over too. Do you equals the X? So that's takes care of our use of going back to the inner rule we have won over. So here the first thing we should do is just replace this radical with you. That's just by definition here and buyer you sub so that just become see you down there. But we also have this X to deal with and the way to get the X is to just go to are you serve in software X, so subtract the one divide by four you get your ex and so that role will replace X. So we have use where minus one over four in place of X and then finally we have R D X and we already found that that's just you over to to you. Now let's go in and simplify this first and see that you can cancel those. This four will go to the top because it's in the denominator within the denominator. So that's a four on time to divide it by the two. Here, you just get it too. And after cancelling the use or just left over with one over, you squared minus one. Okay, now the next step here it. Depending on how you want to solve this, you may have memorized it. You could use the tricks up, u equals one time seeking of data. Or you could just go ahead and spattering and then set up your partial fractions. So here you have the multiply both sides by the denominator on the left, and then you will need to find A and B what's Air beer found? We go ahead and evaluate the integral this term over here instead of the one that's given. So here we find that a equals negative one half the equals one half. So going to the next page, there's our A. There's our b cancel the twos with the one half for the first interview. We just get negative. Natural log, absolute value. You plus one for the second natural log. Absolute value, You minus one. And now, since we've integrated good timeto ad in our seat and then finally we go back to the definition of the use of through this problem to replace all of the EU's in terms of X. If you want, you could drop the absolute value here because the square root plus one is always positive. You see her for about the natural log for the second term, Let me let me take a few steps back ln and then here just as well. Oh, for that this problem only makes sense of excess diggers and zero because if you go to the original and a girl in this problem, there was the X in the bottom by itself. So X can be zero for this problem to make sense. So this term will never be one because X is bigger than zero. So this means that the square root is bigger than one. So when we subtract one, it's still positive here. If you don't want to worry about that. Just keep that slew value there if you're unsure. But in this problem, if you want, you can admit both of the absolute values, and that's your final answer.