Evaluate the indefinite integral.∫tt−2dt\int_{}^{}\frac{t}{\sqrt{t-2}}dt∫t−2tdt

23(t−2)32+4(t−2)12+C\frac23\left(t-2\$$right)^{\frac32}+4$$\left(t-2\$$right)^{\frac12}+C$$

Evaluate the indefinite integral. ∫ t t − 2 d t \int_{}^{}\frac{t}{\sqrt{t-2}}dt ∫ t − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> t d t

we're asked to evaluate the indefinite integral of t over the square root of t minus two dt. So let's go ahead and get started here with our steps for substitution by choosing u and finding du. Now remember that u will often be inside of parentheses under a radical or something like that. We can see that we have this t minus two under our radical, so that's a good indicator that that's what we should choose to be u, that t minus two. If u is equal to t minus two, that then makes du just equal to dt because the derivative of t minus two is just one. Multiplying that by dt is just dt. So with that u and du, we know that we can go ahead and proceed to step two, rewriting our integral only in terms of this u and du. So if t minus two is u and then du is that dt, what do we do about this extra t here? Well, remember that we can go ahead and rewrite whatever variable we're working with, whether that's x or something else like t, in terms of u in order to get our entire integral in terms of u and du. So looking at this here, since u is equal to t minus two, if I add two to both sides here, that then makes t equal to u plus two, just adding two to both sides. So if t is equal to that u plus two, I can replace that in my integral so that t is u plus two. And this allows us to rewrite this integral as u plus two over the square root of u du. Now this is entirely in terms of u and DU, so we can proceed on to step three, integrating with respect to u. Now before we start finding our antiderivatives of our individual terms here, it's going to be helpful to rewrite this in a slightly different way. We know that the square root of u is the same thing as u to the power of one half and since this is a single term on the bottom of a fraction and in the numerator I have two separate terms, I can go ahead and separate this fraction out. So I can rewrite this as u over u to the power of one half plus two over u to the power of one half, just splitting up that fraction. And here keeping that du. Now with this, I can do some simplification because if I have u to the power of one on top and u to the power of one half on the bottom, that really simplifies to u to the power of one minus one half. So that makes this first term u to the power of positive one half because one minus a half is a half. Then for this next term here, I can just write this as two times u to the power of negative one half just so that I don't have to worry about thinking about that as a fraction. So integrating here with respect to you, we wanna find our antiderivatives here. We can do that using the power rule pretty easily now that we've rewritten this. So if I take that one half, I add one. I'm dividing by that new how new power, one half plus one. One half plus one is three halves. If I'm dividing by three halves, I'm really just multiplying by two thirds. So this is two thirds u to the power of three halves. Then plus two, keeping that constant out front, and then adding this adding one to this power, dividing by that new power. Negative one half plus one is positive one half. If I'm dividing by positive one half, I'm really just multiplying by two. So multiplying by two here, that's two times u to the power of one half plus my constant of integration at c. So I'm gonna go ahead and simplify here. Two times two is four. So just rewriting that altogether. And now I wanna go ahead and proceed to that final step here which is replacing u with that original function of x or in this case t. So that original function here was t minus two. We want to go ahead and replace those u's with that t minus two. So this is two thirds times t minus two to the power of three halves and then plus four times t minus two to the power of one half plus that constant of integration c, and this is my final answer here.