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Problem 53 Easy Difficulty

Evaluate the indefinite integral as an infinite series.

$ \int \sqrt {1 + x^3} dx $


$C+\Sigma_{n=0}^{\infty} \left( \begin{array}{c}{1 / 2} \\ {n}\end{array}\right) \frac{x^{3 n+1}}{3 n+1}$

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Video Transcript

so in this probably want to calculate an indefinite integral, um, and we'll be using, um, the binomial serum to help us solve this. So we have is the indefinite integral of the square root of one plus X cubed the X. So with that, we want to use the binomial theorem, which essentially states one plus X to the K is equal to the summation from an equal zero to infinity of Okay, um, for the lack of having being able to write this will just UK over x x two then this is not division. Just to be clear, it's just k over the X. Then we know that X is going to be X cubed, and K is going to be one half. So what will end up getting as a result is one plus one plus X cubed, being equal to one plus x cubed half. And this is the square root of one plus X cubed to be clear. Yeah, we know that these two are equal to each other. So the way that we're going to write this in terms of the summation is that it's equal to the sum from zero to infinity of Okay, one half over and again. That's just that's not supposed to be a fraction times X cubed. And with this in mind, um, what we have is X cubed to the end when we are now, take the integral of this the ax. We end up getting that. This is the summation from end to infinity of one half over and, um, of X to the three and plus one over three. The an plus one. I see. And that will be our final answer. Um, just the main thing to look out for him be careful of is to make sure that there's no fraction bar right here. Um, that's just one half on the top and and on the bottom.