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Numerade Educator



Problem 5 Easy Difficulty

Find the general indefinite integral.

$ \displaystyle \int (x^{1.3} + 7x^{2.5}) \, dx $


$\frac{x^{2} \cdot 3}{2 \cdot 3}+2 x^{3 \cdot 5}+c$


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Video Transcript

in this problem we want to uh find that indefinite integral. Um that you see here, we're going to use this rule for indefinite integral. The integral of X T D N D X is equal to X To the end plus one power All divided by N Plus one. So if you want to integrate X to the end, whatever end is uh the answer will be X city and plus one over N plus one. Now here we're integrating a sum and were allowed to take the some of the integral. Uh so we're going to take the integral of X to the 1.3. D X plus uh Seven plus the integral of seven uh times X to the 2.5 D X. The integral of seven. Uh Well, let's just write it integral of seven Times X to the 2.5 DX. So the integral of a sum equals to some of the integral. All right. So the integral exit at 1.3 DX. We're going to use this rule right here. So the integral exit at 1.3 and is 1.3. So N plus one will be 1.3 plus one or 2.3. So the integral of X 2 to 1.3 Is X to the 2.3 Over 2.3 plus. Now, the integral of seven times a function uh is going to equal seven times the integral, that function. Uh So we can take the seven out of the integral and basically just integrate X to the 2.5. D X which once again, This time end is 2.5. So we're going to use this rule once again, So The integral of X to the 2.5 is going to be X 2-3.5 Over 3.5. And of course when you haven't uh indefinite integral, there's always some constancy uh that we would add at the end. Uh We are done but we can do a nice little simplification. Simplification reflect Final answer. X to the 2.3 power divided by 2.3 plus. Uh this is what works out pretty nicely. 3.5 is half of seven, so seven divided by the 3.5 is to uh so this comes out to be two times x to the 3.5 power. And then of course, plus our constancy. And ah this is our answer. And just so you know, you have a little desk and we're here, let's put that back. Okay, so that is our final answer. That is the indefinite integral