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Evaluate the integral.
$ \displaystyle \int \frac{1}{x \sqrt{4x^2 + 1}}\ dx $
$\ln \left|\frac{\sqrt{4 x^{2}+1}}{2 x}-\frac{1}{2 x}\right|+C$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 5
Strategy for Integration
Integration Techniques
Campbell University
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Let's start off here by doing the u sub. Let's take you to just be x square, then d u equals two x dx and we can solve for X here, so X equals square you. So to rule u d x and then go ahead and solve this equation for DX. You have d x equals d you over to radical you. Now let's rewrite this integral. So we have a one over and then first thing we see is the X Well, we already rewritten X over here. So let's just write this as you to the one half and then instantly have a radical so we can just rewrite. This is the square root of four. You plus one. Yeah. And finally DX. Yeah. Do you, over two years to the one half? Yeah. And let's simplify this. We could pull out this, too, and then combine those you to those radical use in the bottom to get one over you times for you. Plus one. Uh huh. So a different looking in our girl. Little simpler looking in a girl. Let's take another use up here this time V equals for you plus one in the radical. The D V is two over the radical. So this is to over v d u and then solve this for, do you? So we just solved this equation up here for do you? And then you get this V over to D V Also, do you see this YouTube out here outside the radical? So let's solve this for you. So we'll get U equals V squared minus one all over four. So we have one half integral one over, and then the u we just found that over here and then by definition of V, this is just the right here. So times V and then we have our d you at the very end. And that's just V over to D. V right now. Before we had a great let's simplify. We have a four here on the bottom, but this four will end up at the numerator. So those cancel and we see that these vis cancel. So we just have the integral of the square minus one. And the first thing you should do here is just factor that before you try the partial fractions, of course, you could also do a tricks up here. If you wanted to, that would work. So this is the alternative. So here, if you use partial fractions, you'll end up with the composition of the forum. A over B minus one B overview plus one. So you have to go ahead and find a and B. And when you do so you're getting one half or a and then minus one half for me. So I just factor out that mine is there and then integrate these natural algorithms. Don't forget the absolute value. And also don't forget the plus. See at the very end. Now we're almost finished. Let's go to the next page and just rewrite what we previously had so here. V minus one minus one half l n V Plus one plus e and then V by definition radical of for you plus one. So let's replace all the vis with the radical and then recall the definition of you are first substitution for the original problem. So go ahead and replace the use with X squared if you want. Here you can drop the absolute value because the expression inside is positive. But if you're unsure, just go ahead and leave that absolute value there. Oops. So with that, plus ones inside the radical here and then we have another plus one on the outside. Add that constant of integration, See? Yeah, and that will be our final answer.
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