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JH
Numerade Educator

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Problem 54 Medium Difficulty

Use integration by parts, together with the techniques of this section, to evaluate the integral.

$ \displaystyle \int x \tan^{-1} x\ dx $

Answer

$\frac{\left(x^{2}+1\right) \tan ^{-1} x}{2}-\frac{x}{2}+C$

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Video Transcript

let's use integration by parts, possibly with the techniques of the section. We may have to do partial fractions after immigration My parts well issues. This is our starting point here and maybe easiest to just go ahead and take you to be our camp. We know the derivative of that one over X squared plus one D x devi is X tx So then the expert over too. So you think in aggression my parts recall that this is UV minus in roll video. So for you, times v ex cleared over two times are tan and then minus in a girl And then we have VD you. Let's go ahead and pull that one half are the rules. Now before we try partial fraction to composition if it's even necessary First we should do long division here because the numerator has degree That's equal. So the denominator Saagar decide to do that and we have a remainder of minus one. This means we can rewrite this all. So here we have our quotient which was one and there are remainder was minus one. So we have minus one over X squared plus one. Now we know from basically going backwards from to you to you, We know that the integral of one over X squared plus one is our ten. So otherwise, if you have forgotten that factor didn't see this up here, you could go ahead and do it shrinks up if you had to. And that will give you the anti derivative. So now let's evaluate X squared over to our ten and then we have minus one half and then x minus Ark ten of X. We've just evaluated the integral. So don't forget to add that constant of Integration City. And then the last step is maybe we can go ahead and just distribute this net negative one half. So I'll go to the next page to write that X squared over to ten in verse of X. This was our UV term, and then the general gave us this and that's your final answer