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Let's say that we have the indefinite integral,

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and the function is 3x^2 plus 2x times e to the x^3 plus x^2 dx.

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So how would we go about solving this? So first when you look at it,

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it seems like a really complicated integral; we have this polynomial right over here

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being multiplied by this exponential expression and over here, the exponent,

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we essentially have another polynomial. It seems kind of crazy.

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And the key intuition here, the key inside is that you might want to use a technique here called

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u-substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution,

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and over time, you might even be able to do this type of thing in your head.

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U-substitution is essentially unwinding the chain rule. In the chain rule --

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I'll go in more depth in another video where I really talk about that intuition.

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But the way I'd think about it is, well I have this crazy exponent right over here,

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I have x^3 plus x^2. And this thing right over here happens to be the derivative of

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x^3 plus x^2. The derivative of x^3 is 3x^2; derivative of x^2 is 2x,

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which is a huge clue to me that I could use u-substitution.

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So what I do here is, this thing where this little expression here where I also see its derivative

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being multiplied, I can set that equal to u. So I can say, "u is equal to x^3 plus x^2."

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Now what is going to be the derivative of u with respect to x?

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du/dx (well, we've done this multiple times) is going to be 3x^2 plus 2x,

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and now we can write this in differential form. And du/dx isn't really a fraction

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of differential of u divided by differential of x, it really is a form of notation.

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But it is often useful to kind of pretend it's a fraction.

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And you could kind of view this, if you want to just get a du,

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if you just want this in differential form over here, how much does u change for given change in x,

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you can multiply both sides times dx. And so if we were to pretend it's a fraction,

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and it will give you the correct differential form,

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you're going to be left with du is equal to 3x^2 plus 2x dx.

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Now why is this over here, why did I go through the trouble of doing that?

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Well we see we have a 3x^2 plus 2x, and it's being multiplied by a dx right over here.

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I could rewrite this original integral as the integral of -- let me do that in that color --

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of 3x^2 plus 2x times dx times e -- let me do that in that other color --

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times e to the x^3 plus x^2. Now, what's interesting about this,

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well this stuff that I have in magenta here is exactly equal to du.

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And then this stuff I have up here, x^3 plus x^2, that is what I said what u equaled to.

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That is going to be equal to u. So I could rewrite my entire integral --

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and now you might recognize why this might simplify things a good bit --

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it's going to be equal to, and what I'm going to do is I'm going to change the order.

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I'm going to put the du, this entire du, I'm going to stick it on the other side here

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so it looks like more of the standard form that we're used to seeing in our indefinite integrals in,

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so it's going to be, we're going to have our du times e to the u.

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And so what would the anti-derivative of this be in terms of u?

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Well, the derivative of e^u is e^u; the anti-derivative of e^u is e^u.

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So it's going to be equal to e^u, now there's a possibility that there's some type of constant

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factor here, so let me write that. So plus C. And now, to get it in terms of x,

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we just have to un-substitute the u. We know what u is equal to.

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So we could say that this is going to be equal to e -- instead of writing u,

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we could say u is x^3 plus x^2. And then we have our plus C. And we are done!

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We've found the anti-derivative. And I encourage you to take the derivative of this,

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and I think you will find yourself using the chain rule and getting right back to what we had over here.