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## Integrating with u-substitution

# 𝘶-substitution: rational function

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

## Video transcript

So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx. So how can we tackle this? It seems like a hairy integral. Now the key inside here is to realize you have this expression x^4 + 7 and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3. Derivative of x^4 is 4x^3; 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- -- I'll write the whole thing. U-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 'cause it can figure out this part and 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^4 plus 7. Now, what is du going to be equal to? du, I'm doing it in magenta. du, well it's just going to be the derivative of x^4 plus 7 with respect to x, so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- and I'll just rewrite it up here so that it becomes very obvious; our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? 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 U's -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!