Evaluate the integral. ∫−6x2+3x+5x3−xdx\int_{}^{}\frac{-6x^2+3x+5}{x^3-x}dx

Evaluate the integral. ∫ − 6 x 2 + 3 x + 5 x 3 − x d x \int_{}^{}\frac{-6x^2+3x+5}{x^3-x}dx

in this video, we are asked to evaluate the following integral, which is the integral of negative 6x squared plus 3x plus five over x cubed minus x, and that's all integrated with respect to x. Now, how do we go about solving this problem? Well, what I need to do is take a look at the rational function that I'm dealing with here and think about general strategies for solving this. Now you could try a variable substitution down here. You could also try a variable substitution on the numerator. But no matter how you did a variable substitution, you'd find that everything inside the integral is not going to simplify by taking that strategy. So we're gonna need to do something different. Now I think the best way to approach this is gonna be to try to get some form of partial fractions. But let's actually first see how this would look. So what I'm going to do is recognize something. We can see that the degree in the numerator is less than the degree in the denominator, x squared versus x cubed. So we don't need to perform polynomial division here. So since we have a lesser degree in the numerator, we can skip the polynomial division step. What I do want to try and do is factor the denominator. So we're going to get negative 6x squared plus 3x plus five. We'll keep this the same on top. Then I'm going to divide this by a factored version of this. Now the way I can do this is by recognizing something. I can take an x, and I can factor it out here. So factoring out an x, we get x times x squared minus one. So that's what we get in the denominator. Now, at this point, there actually is a way I can factor this farther. Because notice x squared minus one, we could also write that as x squared minus one squared. And the reason I'm writing it like this is because this is the difference between squares. One squared is the same thing. It's just one. But if I go ahead and rewrite this all out, so keeping this integral the same, we'll have six x squared plus three x plus five on top divided by and then I can write this as a difference of squares, which we know that the difference of squares will factor to x plus one times x minus one. So now we have the difference of squares written out, and notice how everything is split into distinct linear factors in the denominator. So I can go ahead and apply that and use some partial fraction decomposition. From there, I can integrate each partial fraction separately, and that will allow me to get to a final solution. But let's go ahead and do that decomposition step first. Now this is oftentimes what takes the longest amount of time is just trying to figure out how we can decompose this fraction, but we should hopefully have the skills to be able to do this at this point. So doing this, I can see we have three distinct linear factors in the denominator. So I'm going to rewrite this as some constant a over this factor x. We'll have plus some constant b over this next distinct linear factor x plus one. Then we'll have the constant c over this distinct linear factor x minus one. Now we've split this up. Now our next step from here is gonna take both sides of this equation and multiply this by the common denominator, which is this piece right here, x times x plus one times x minus one. And I need to distribute this to each place inside of the brackets. Now we'll go ahead and start over here. If I distribute this common denominator over here, notice that the common denominators are going to cancel completely. That's the same thing. So multiplying that out, all I'm going to have left on the left side here is 6x squared plus 3x plus five. Now what do I do from here? Well next I need to move to this next factor, which has the a. And notice if I multiply this out right here, I can get the x's to cancel completely. So that's going to leave me with a, then we'll have times x plus one times x minus one. Now next, I'll move to this b. Notice that if I multiply out the common denominator here, the x plus one will cancel right about there. So I'm going to have b, and then we said x plus one cancels, so we're just gonna have x times x minus one. Now let's move to the c. For the c, I can multiply this into there, the common denominator, and notice multiplying this out, the x minus ones cancel completely. So I'm gonna be left with here is c times x times x plus one. So now we have this equation written out. Now from here, you have a couple of different options. You could either try to create a system of equations, or you can do strategic substitutions for x to try and solve for your constants, a, b, and c. I think I'm gonna use strategic substitutions for x because I think that's gonna be a little quicker, and that allows us to keep it in this form without having to distribute and foil everything out. So what I'm first going to do is let x equal negative one. Now the reason that I'm doing that is because notice if we had a negative one right there, it would cancel with that positive one, setting this whole factor to zero. And same over here. A negative one right there would cancel with the positive one, setting that whole factor to zero. It is helpful to try and eliminate factors when doing these problems. So negative one for x is a great candidate here. So let's go ahead and do that. So we're going to have six times negative one squared plus three times negative one plus five on the left side here, replacing x's with negative one. Then we'll have a times negative one plus one times negative one minus one. Then we'll have plus b times negative one times negative one minus one, plus c times negative one, and it's going to be times negative one plus one. Negative one plus one is zero in both of these cases, So both of these factors are going to go to zero. So what we're gonna be left with is six times negative one squared, that's just gonna be six, plus three times negative one, which will give us a minus three, plus five, and that's going to be equal to in this case, we'll just have b times negative one times negative one minus one. Well, negative one minus one is negative two times negative one. That gives us two b. So we'll have six minus three, which is going to be three plus five, so we'll have eight is equal to two b. And I can divide this two on both sides of the equation here, giving us that b is equal to eight over two, which is four. So now we've gone ahead and solved for b. And if b in this case is four, I can replace it in both these equations right here. So that goes ahead and shows us that we've solved for b, but now we need to find a and c. What can I do from here? Well, next, what I can do is I can let x equal positive one this time. Now why am I doing that? Well, the reason I'm letting x equal positive one is notice if x was equal to one, one minus one would give us zero, and it would be the same for over here. So I can eliminate two of these factors once again. So we'll have six times one squared plus three times one plus five is equal to in this case, we'll have a times in this case, it's one. So one plus one times one minus one plus four times one times one minus one plus c times one times one plus one, just replacing every x with one. So we'll have six plus three, which is nine, plus five on the left side, which all comes out to 14, and then we're going to have this whole factor right here is going to go away since the one minus one is zero. And then this whole factor is gonna go away since the one minus one is zero. So we're just going to have c times one times one plus one, which is the same thing as two c. Divide that two on both sides of the equation, canceling the twos there, giving us that c is 14 over two, which is the same thing as seven. So we've now solved for c by figuring out that c is equal to seven. So I can replace it in the fraction up here, and I can also replace the c right about there. So we've now solved for a, and we've solved or we've now solved for b, and we've now solved for c. And the last thing that we need to find here is this a. How do I find a? Well, it seems that we've gone ahead and set the x's equal to things that would eliminate this piece and that piece, allowing us to eliminate our factors. But there's one more thing that we can set x equal to to eliminate a bunch of factors. And what I'm gonna do is I'm gonna set it over here. What we can do is let x equal zero. Now why am I doing that? We'll notice that if x was zero right here, this whole factor would go away. If x was zero right here, that whole factor would go away. So what we can do is set x equal to zero, giving us six times zero squared plus three times zero plus five is equal to, in this case, is going to be a times zero plus one times zero minus one. And then at this point, I already know these two factors are gonna cancel, so I'm just gonna write this as plus four times zero times zero minus one plus seven times zero times zero plus one. Anything times zero is just zero, so that goes away, that goes away, and then this goes away, and this goes away. So what are we gonna be left with? We'll be left with five on the left side, and then on the right side, we'll have a times one times negative one, which will give us negative a, meaning that a is simply equal to negative five. So going up here, we'll get that our a is negative five. So now that we've gone ahead and found a, and we found b, and we found c, we have now found all of these constants that we were looking to solve for. So this right here would give us our broken down rational function, and we've gone ahead and split this into partial fractions. Now this is not the final solution here because this was not a partial fraction decomposition problem. This was an evaluate the integral problem. But by going through this entire process of doing these partial fractions, we have now set up something that will allow us to solve the remainder of this problem. Notice we have now set up a case where we now know this entire rational function that we started with is equivalent to the integral of, in this case, it's negative five over x plus four over x plus one plus seven over x minus one, and we're integrating this whole thing with respect to x. Now that we've broken this down in such a fashion, we can actually see how to individually integrate each one of these pieces, which we are allowed to do. That's a rule for integrals that we can apply. So since we broke this down to partial fractions, this integral becomes a lot more simple. So what I can do is recognize if we integrated negative five over x, that would just give us negative five times the natural log of the absolute value of x. Right? We know that integrating one over x gives us that, and then this negative five is just a constant that gets pulled to the outside of that integral. Now if I integrate this piece right here, well, that four would come on the outside, and then integrating one over x plus one would give us plus four times the natural log of x plus one. And then we can go ahead and integrate this piece, which would give us that constant seven times the integral of one over x minus one, which comes out to the natural log of the absolute value of x minus one plus the constant of integration c, and there is the solution to the problem. So that is how you can use partial fraction decomposition to integrate complicated rational functions. Hope you found this video helpful.

Evaluate the integral. ∫ − 6 x 2 + 3 x + 5 x 3 − x d x \int_{}^{}\frac{-6x^2+3x+5}{x^3-x}dx

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

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Multiple Choice

Evaluate the integral.
$\int_{}^{} \frac{- 6 x^{2} + 3 x + 5}{x^{3} - x} d x$

$- 6 \ln ∣ x ∣ + 3 \ln ∣ x + 1 ∣ + 5 \ln ∣ x - 1 ∣ + C$

$- 5 \ln ∣ x ∣ + 4 \ln ∣ x + 1 ∣ + 7 \ln ∣ x - 1 ∣ + C$

$- 5 \ln ∣ x ∣ + 4 \ln ∣ x - 1 ∣ + 7 \ln ∣ x - 1 ∣ + C$

$- 6 \ln ∣ x ∣ + 3 \ln ∣ x - 1 ∣ + 5 \ln ∣ x + 1 ∣ + C$

Verified step by step guidance

Rewrite the integrand by factoring the denominator. The denominator x^3 - x can be factored as x(x^2 - 1), which further factors into x(x - 1)(x + 1). This gives the integrand as (-6x^2 + 3x + 5) / [x(x - 1)(x + 1)].

Decompose the fraction into partial fractions. Write (-6x^2 + 3x + 5) / [x(x - 1)(x + 1)] as A/x + B/(x - 1) + C/(x + 1), where A, B, and C are constants to be determined.

Multiply through by the denominator x(x - 1)(x + 1) to eliminate the fractions. This results in the equation -6x^2 + 3x + 5 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1). Expand and collect like terms to solve for A, B, and C.

Once A, B, and C are determined, rewrite the integral as the sum of three simpler integrals: ∫(A/x) dx + ∫(B/(x - 1)) dx + ∫(C/(x + 1)) dx. Each of these integrals can be solved using the natural logarithm rule: ∫(1/u) du = ln|u| + C.

Combine the results of the three integrals into a single expression, simplifying the coefficients of the logarithmic terms. Ensure the final answer is in the form -5ln|x| + 4ln|x + 1| + 7ln|x - 1| + C.

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