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Problem 53 Hard Difficulty

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

$ \displaystyle \int \ln (x^2 - x + 2)\ dx $


$$\left(x-\frac{1}{2}\right) \ln \left(x^{2}-x+2\right)-2 x+\sqrt{7} \tan ^{-1} \frac{2 x-1}{\sqrt{7}}+C$$


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Video Transcript

let's use integration by parts to evaluate this integral. So here's our choice for you. And then take the derivative to get to you. That's using the chain rule there. And we're left over with DV equals DX which gives V equals X. So, using the formula, you see minus integral vdo Jim, we have X ellen X square minus X plus two that's UV and then minus integral being to you. And so now we have an integral to evaluate in here. This is the and a girl feet you are. So since these air both degree, too, we should do long division here. So here we get this expression here. So this is from doing polynomial division. Now, let's go ahead and rewrite this source code on the next page. And then we're subtracting this integral and then our growers attracting a school head and in that denominator completely square no. So here I'll just rewrite this so that first in a girl just plotted to x to the ex and then x minus I have. And then plus seven halves I did was rewrite this. What's really shit is a be a positive for with the double negative there I have a double minus. That's positive. One half and then seven haps is for so this and this correspond to that? No. Now let's go ahead and rewrite this. So evaluate that first, Integral for the second hand, our girl here, you can go ahead and do a U substitution u equals X minus a half. So they're after doing the use of we should get We had you still have this minus one half natural log X squared minus X plus two. And then for this final integral over here. If we do drinks up X minus one half equals room seven over to Tan data. It's a working on that integral we end up with Plus Root sent it and then ten in verse to explain this one over Route seven. And then now the last things such as going on and that constant of integration. See, And that's your final answer