U Substitution Example 6 Hi everyone. Welcome back. We’re going to do another U Substitution problem today. But first of all, the integral of 5x/5+2x2dx so, U Substitution remember, we’re looking for something to assign U two in our problem that will when we take the two it make a little bit make our problem easier to solve. In this case is kind of give away because the denominator, it’s so much more complicated than the numerator so, we’re going to add and try some U to the bottom. So, we’re going to say that U is 5+2x2 and we’re going to take the derivative of U so again, we call that du and the derivative of this, the 5 goes away and 2x2, we will multiply 2 the exponent by the coefficient so, two times two is four, then we keep the x and then of course we subtract 1 from the exponent so 2-1 is 1. So you could write 1 there, but of course it’s implied to be on that. And derivative therefore is 4x so, now that we taken the derivative, I always say, remember to add dx because we have to this equation 4dx so that we can plug our answer back in for this dx up here. So, we’re going to ahead and solve for dx and the way that we do that is by dividing both sides by 4x so, I will cancel on this side and we’’ end up with dx=du/4x. So now that we have our U and we solved for dx, we can go ahead and plug these two things back into our equation here. So, what does it look like when we plug back in, is the integral of – of course 5x on the top stay the same, we did not touch that. But then, we assigned u to 5+2x2 so, we’re going to have U on the bottom here and then dx – we solved for dx down here and we got du/4x so, we plug in du/4x for dx so, we end up with this. So, we got 5x/u times du/4x, which is great because now on our situation here, we can cancel out this xs. You see we have x in the numerator and then x in the denominator so, we can cancel those out. We can also take the 5/4 out in front of the integral, just simplify. Whenever you have one term, everything here is multiply together so, this is one term and you have a coefficient on that term, 5/4 here. You can pull that out in front. So we ends up looking right is this. We pulled the 5/4 in front and then we have the integral. The xs here canceled. They go away. And we end up with (1/u)du, which is perfect because this is now something that we can easily take the integral of initiative word that you should memorize or ideal you’ll memorize so that you’re quick, but you can plug it in your calculator and have that, and test if you need it. But the integral of 1/x is the natural log for ln of the absolute value of x. So, when we take the integral with this, this going to look like – we’ve got our 5/4 in front and now we’re taking the integrals so I just go ahead and draw a big parenthesis because we have 5/4 is multiply by everything that where were going to take the integral off. So, like what I was saying, natural log of the absolute value of x. So in this case, we have 1/u so, we’re going to have natural log or ln of the absolute value of u. These two lines here simplify absolute value; it does not mean that if you end up with like negative number in here or if you put down same u as representing negative five and you plug in negative five there, this absolute value indicated that because its positive five always. Absolute value means that if there’s negative number, we trade this positive number. So anyway, just as a rule, this is a formula that we’ve used. This is our integral and then do not forget we always have to add C to account for the constant and then the only thing we have to do here is plug in for you – remember we assigned you to 5+2x2 so, our final answer is actually 5/4 ln¦5+2x2¦+C. And that’s it. See you next time.