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Current time:0:00Total duration:4:04

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

so we want to take the indefinite integral of 4x to the 3rd over X to the 4th plus 7 DX so how can we tackle this this seems like a hairy integral now the key inside here is to realize that you have this expression X to the 4th plus 7 and you also have its derivative up here the derivative of X to the 4th plus 7 is equal to 4x so third derivative of X to the fourth is 4x to the third derivative of 7 is just 0 so that's a big clue that u substitution might be the tool of choice here u sub I'll just write u well write don't they use substitution could be the tool of choice so given that what would you want to set your u equal to and I'll let you think about that because if you can figure out this part then the rest will just boil down to a fairly straightforward integral well you want to set u be equal to the expression that you have its derivative laying around so we could set u equal to X to the fourth X to the fourth plus seven now what is d u going to be equal to D u I will do it in magenta D u well is just going to just be the derivative of x to the fourth plus seven with respect to X so 4x to the 3rd plus 0 times DX times DX I wrote it in differential form right over here but it's a completely equivalent statement to saying that D U the derivative of u with respect to X is equal to 4x third for X to the third power when someone writes D u over DX like this this is really a notation to say the derivative of U with respect to X it really isn't a fraction in in a very formal way but oftentimes you can you can kind of pseudo manipulate them like fractions so if you wanted to go from here to there you could you could kind of pretend that you're multiplying both sides by DX but these are equivalent statements and we want to get it in a differential form in order to do proper u substitution and the reason why this is useful and I'll just rewrite it up here so that it becomes pretty obvious our original integral we can rewrite as 4x to the third DX over X to the fourth plus seven and then it's pretty clear what's D you and what's you you which we set to be equal to X to the fourth plus seven and then D U is equal to this it's equal to four X to the third DX we saw it right over here so we could rewrite this integral and I'll try to stay consistent with the colors as the indefinite integral what we have in magenta right over here that's D u D u over try to stay true to the colors over X to the fourth plus seven which is just u or we could rewrite this entire thing as the integral of 1 over u 1 over u D u 1 over u D u well what is the indefinite integral of 1 over u D u well that's just going to be equal to the natural log of the absolute value and we use the absolute value so it'll be defined even for negative use and it actually does work out and another video where I show you that definitely does the natural log of the absolute value of U and then we might have had a constant there that law was lost when we took the derivative so that's what we that's essentially our answer in terms of U but now we need to unsubstituted when we unsubstituted this is going to be equal to the natural log of the absolute value of well U is X to the 4th plus 7 X to the 4th plus 7 not C plus 7 and then we can't forget our plus C out here and we are done

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