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Numerade Educator

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Problem 12 Medium Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int x \sqrt{2 + x^4}\ dx $

Answer

$\frac{x^{2}}{4} \sqrt{2+x^{4}}+\frac{1}{2} \ln \left(x^{2}+\sqrt{2+x^{4}}\right)+C$

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Video Transcript

Okay. This question wants us to integrate this function using a table. But the issue is that this doesn't look like any of the forms we know. So we're gonna have to put it into that for so a lot of square root expressions have a use squared in the room. So let's pick you squared equal the exit fourth. And this implies that you Cols X squared or squared of you is equal Dax. So from there we confined that d x is equal to one over to root you. Do you? So now we can plug everything back in so we have the integral of X is squared of you Times x to the fourth That's you squared. We're leaving the two alone divided by to root you do you? Because we have to have that factor. And now we see we get a nice cancellation, doesn't root use and we're left with 1/2 times the integral of square root of two plus u squared. Do you? And this is something that we know the answer based on our table. We see that this is you over too. Time squared of two plus u squared, plus the natural law of you plus square root of you squared plus two plus c, of course. And no, all we gotta do is back. Substitute everything in So this is equal to well, you is equal to X squared. So we get distributing X squared over four times the square root of X to the fourth plus two because you squared is extra fourth plus 1/2 times the natural log of X squared plus the square root of two. Sorry for consistency will write it this way, plus C and this will be our final answer making sure we substituted back in properly.