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JH
Numerade Educator

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Problem 34 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{x^5 + x - 1}{x^3 + 1}\ dx $

Answer

$$
\frac{x^{3}}{3}-\ln |x+1|+C
$$

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Video Transcript

Let's evaluate the given integral since the numerator has degree larger than the denominator. Let's go ahead and do polynomial division exit the fifth over X cubed. That's ex clear, so it's couldn't multiply that out. Then you subtract and your leftover with negative X squared plus X minus one, and this will be a remainder because X cubed is not going to x squared. So coming back to the original problem, we can rewrite this integral. Our quotient was X squared. That's from up here. And then we had our remainder. So let me go ahead and pull up this minus sign. So we have a minus. Actually, like, I am going to take a step back here. I'm under just right. This is a plus plus our remainder x cubed plus one. Okay. And then now break this into two into girls. And for the second inaugural, let me just plug this minus and from the X Square. So then I'll have X square and then we have to change the sign and the other terms because we factor out a minus sign. Yeah, there's a formula for X Q plus one. This is coming from This is an expression of the form a queue. Plus be cute. We're here Is X B is one. And there's a formula that says this is equal to a plus speed a square minus a B plus peace Where so, using this formula, we have X plus one. But then we have X squared minus X plus one, and we could see that there's a cancellation here. This quadratic terms cancels nicely with the other. So we're just left with in a girl. It's square minus. So she put a DX. They're technically minus in a girl, one over X plus one. Now, for the first interval, we can just use the power rule. That's X cubed over three for the second. And the girl that, plus one is bothering you. Feel free to go. Do you, sir? Here. You could take you to just be f plus one. Then you have natural log. Absolute value X plus one. And then don't forget the constancy of integration. And that's our answer.