Use the method of partial fractions to evaluate the integral.∫5x3−x2+7x−2(x2+1)(x2+2)dx\int_{}^{}\frac{5x^3-x^2+7x-2}{(x^2+1)(x^2+2)}dx

ln ⁡ ∣ x 2 + 1 ∣ − t a n − 1 ( x ) + 3 2 ln ⁡ ∣ x 2 + 2 ∣ + C \ln\left|x^2+1\right|-tan^{-1}\left(x\right)+\frac32\ln\left|x^2+2\right|+C

Use the method of partial fractions to evaluate the integral. ∫ 5 x 3 − x 2 + 7 x − 2 ( x 2 + 1 ) ( x 2 + 2 ) d x \int_{}^{}\frac{5x^3-x^2+7x-2}{(x^2+1)(x^2+2)}dx

in this problem, we're asked to use the method of partial fractions to evaluate the following integral. Now since this problem is specifically asking us to use partial fractions, we can already tell that a variable substitution is not going to be the method we use to solve this problem, whether it be on the top or one of these pieces on the bottom. Now, you don't you can always just check to see if the variable substitution is going to work, but in this case, it won't. So this is the situation where we will need to use the partial fractions to evaluate the integral. Now how do I go about doing that? Well, looking at this integral, I can see we have a pretty long polynomial in the numerator, so it may be tempting to think that we're gonna need to do long division first, But if you look at the denominator here, notice we have these two pieces. And if we were to go ahead and use the foil method to group these together to multiply everything, well, notice we would end up having to multiply x squared by x squared, which would give us x to the fourth power. X to the fourth power is a higher exponent than x to the third power, so we actually don't have to do long division here with polynomials. So what I can just do is take a look at this fraction that we have, and try my best to break this down. So we have five x cubed minus x squared plus seven x minus two, and all of this is going to be divided by x squared plus one times x squared plus two. So just rewriting out just this fraction by itself, so I can do these partial fractions a little bit easier. Now what I need to do is take a look at what we have in the denominator right here. Notice this is an irreducible quadratic factor. We have a quadratic factor there, and there's no way of factoring this or breaking this down, so that's an irreducible quadratic factor. So we'll first have a x plus b, a linear equation, over that factor, x squared plus one. Now notice here we have another irreducible quadratic factor that shows up. So we'll get another linear equation, c x plus d, over that piece, x squared plus two. Now we've gone ahead and we've split this into separate fractions. So from here, I need to find a way to solve for these constants, a, b, c, and b. Now there's a way that I'm going to do this. I'm gonna first take both sides of this equation and multiply it by the common denominator. In this case, that's x squared plus one times x squared plus two. So let's go ahead and multiply this all out. Now multiplying this on the far left side of this equal sign, I can cancel the common denominator entirely. So what that's going to give me is five x cubed minus x squared, plus seven x minus two, since that denominator I know is entirely going to cancel. Now next I'll go ahead and distribute into this term, and notice here the x squared plus one will cancel completely there. So that's going to give me this whole thing equals a x plus b times, and in this case, the x squared plus one cancels, so it'll be times x squared plus two. Now next, I'll move to this last factor here, the c x plus d term, which I can multiply in here, giving you x squared plus two to cancel entirely. So if the x squared plus two cancels entirely, well, that's going to leave me with c x plus d times, and then we'll just have x squared plus one. Now I've gone ahead and expanded this, so we end up with this equation right about here. Now what can I do to solve the problem from here? Well, this is a case where you could either try to make strategic substitutions for x, or you can try to get a system of equations. Now this is why it's important to know both methods, because oftentimes, it's pretty straightforward to deal with the substitutions for x. However, looking at this equation here, I notice if I try to start making substitution for x, we're going to get a lot of terms and a lot of constants that will still remain. So if I set x equal to zero, that will go away, but it looks like everything else about that will go away too, but it looks like everything else here, the b's and the d's will remain. And this is just gonna be a mess. So I am just gonna go ahead and set up a system of equations to solve for all the constants at once. Now this is gonna be a bit tedious because I am gonna have to use my foil method twice here on these two terms, but we can do this one at a time. So let's start by dealing with the a x plus b times the x squared plus two, that's going to be this whole piece right here. Well, doing this, I can multiply the a x times the x squared, giving me a x cubed, a x times x squared is a x cubed. And now I'm going to have plus, and I'm going to multiply the a x by the two. That's gonna give me two a x. Next, I'm gonna multiply the b by the x squared. That gives me b x squared. And I'm gonna multiply the b by the two, giving me two eight. So that's what happens when we go ahead and foil all of that out. Now next, I'm going to foil this piece out, and this is going to be the same strategy we did right here. So we wanna have c x plus d times x squared plus one. Well, first, I'll multiply the c x by the x squared. I know that's going to give me c x cubed. Then I'll multiply the c x by the one. That's going to give me c x. Now I'm going to multiply the d by the x squared or the yeah. The x squared give me d x squared, and I'm gonna multiply the d by the one giving me d. So that's what happens when we foil everything out right about there. So if I were to rewrite this whole thing, well, we would have this whole polynomial that remains on the left side of the equal sign. This is not going to change. But then all of this is going to be equal to these foiled out expanded equations that we found for each of these terms. So we're gonna get a x cubed plus two a x plus b x squared plus two b, which is this piece right here, plus, and then we're going to move on to this portion, we're going to have c x cubed plus c x plus d x squared plus d, and that's what happens when we go ahead and write this all out. So now that we've written all these terms out, I now want to focus on grouping common terms together. Now this is the part where we will get things that have a similar x raised to the same exponent, and we can group the coefficients in front. Now I see that we have an a x cubed, and I see we also have a c x cubed right about there. Now keep in mind this whole time, this polynomial is just gonna stay the same, we don't need to worry about this just yet. But if I went ahead and group this if I wanna go ahead and group this together, we'll take the a x cubed and the c x cubed, and I'm gonna group those to be a plus c times x cubed. Next, I'm going to move look for x squared terms. I see that I have a b x squared right about there, and I have a d x squared right about here. So because I have a a b x squared and a d x squared, I can group those together as well. This It's gonna give me plus, and then we'll have b plus d x squared. Now let's take a look for just x's. Well, I can see that we have just an x term right there, and just an x term right there. So because of this, I can group this together to be the coefficients in front two a plus the coefficient in front c times x. Now lastly, I'm gonna move to just the constants, and looking at the constants, I can see we have a two b and we have a d. So I'm just gonna put the two b and the d at the end right there. So now we've grouped things together, and next would be equating the coefficients to give us a system of equations where we can solve for all of these constants and variables that we're trying to find. Well, let's go ahead and start by looking at the x cubed term. So I'm gonna take this, everything we have in front of this x cubed, and set it equal to everything in front of the x cubed right there. So that would be a plus c is equal to five, and and we have our first equation right there. Now, let's move to the x squared. Well, I can see that we have this b plus d in front of x squared, and then we have a negative one in front of x squared right there. So, I can say we have b plus d is equal to negative one. You can imagine there's still one because that's one times x squared and then it's negative, so that's what we get there. And now next, I'll take a look for the x's. Well, looking at the x's, I can see that we're going to have these coefficients two a plus c in front of the x, and we have a seven right there. So equating the coefficients, we'll have two a plus c is equal to seven. And then lastly, we'll move to these constants right here. Well, for these constants, you see we have two b plus d, and then we have one constant here, negative two. So we're going to have that two b plus d is equal to negative two. Now we have our system of equations. So how do I go about solving this? Well, this is the part that can oftentimes take a lot of time just because we can see that none of these equations give us a super straightforward case, but I can go ahead and solve this one variable at a time. It looks like the simplest equation we have says that a plus c is equal to five. Now, what I'm going to do here is I'm going to go ahead and solve for c in this equation, and to do that I would just subtract a on both sides of the equation, so that would give me that c is equal to five minuteus a. Just moving that a to the other side make it negative, this would give us our c. Now why did I go ahead and do that? Well if I went ahead and solved for c like this, notice I could substitute a c into that equation. Now why would that be helpful? Well notice if I substituted a c in terms of a, we would only end up with a as the variable in this equation. So going to this equation right down here, I could say we have two a plus, and then c, which is five minuteus a, is equal to seven, and I just need to go ahead and solve for a, and that's going to knock out the a variable. So if I want to find what this a is, well we have two a plus five minuteus a is equal to seven. Now two a minus a is just going to give us a, so if a plus five equals seven, and I can subtract this five on both sides of the equal side, giving me that a is seven minus five, or two. So we have that a is equal to two. Now, now that we've solved for a in this equation though, notice I can go back up to this equation, and I can solve for c. Because if a is equal to two, well, we said that c is equal to five minus a, so c is going to be five minus two. And five minus two is just three, so that means that c is equal to three. So now I've found my a, and I've found my c. So all I'm looking for from here are my b and my d. Well, what I can see here is that we have this equation that has b's and d's, and we have that equation that has b's and d's, so unfortunately, there's no place that we can go ahead and substitute our c's and our a's in. But since I can see that we have this equation, which looks more simple than that one, what I can do is solve for one of these variables. In this case, I'm going to solve for d. So going over here, we are told that b plus d equals negative one. If I wanna solve for d, I'll just subtract that b on both sides of this equation, giving me that d is equal to negative one minus b. So now we've solved for d. So now that I've solved for d, I can go to this last equation, which we haven't touched yet, and substitute this in for d. So we'll have two b plus, and then that's going to be d, which we just solved for right there. So two b plus negative one minus b is equal to negative two. So at this point, we can say two b minus one minus b equals negative two. Two b minus b, that's just gonna give you b. So we have b minus one equals negative two. I can add that one on both sides of the equation, canceling the ones right there giving us that b is equal to negative two plus one, which is negative one. So now we've solved for b. And if we've solved for b, then we have a straightforward way of solving for d. Just take that b and put it into this equation. So we'll have that d is equal to negative one minus, and then we said b was negative one, cancel the negative signs giving us negative one plus one, meaning d is just gonna equal zero in this case. So now we've solved for a, b, c, and d, and I'll just write these up here. We've solved that a is equal to two. We figured out that b is equal to negative one. We figured out that c is equal to three, and then we figured out that d is equal to zero. So that means I can go back up to the fractions that I started with right about here, and and I can replace these variables, these constants, I should say, with what we have found. So the a, I said, was two. We figured out that was two. The b, we figured out was negative one, so this whole thing is going to be minus one. The c, we figured out was three, so I can replace that c with a three. And then the d, we figured out was zero, so I can just remove this d entirely. So now we have these partial fractions split up like so. So going back up here, what I can say is this whole thing is going to be equal to the integral, and in this case, we're going to have two x minus one over x squared plus one plus three x over x squared plus two. So now we are integrating all of this with respect to x. So that is how we can go ahead and break this down. So this is the integral we're now trying to solve. But how do I go about solving this? Well, there are going to be a few steps for dealing with this integral. So I'm gonna go ahead and move this down here just so we have more space. So we've done our partial fraction decomposition, and we've broken it down into the integral of these two simpler rational functions. But let's go ahead and break this down one at a time. So first off, I'm going to need to integrate this, and then I'm going to need to integrate that. Now starting with integrating this function right here, we have the integral of, in this case, is going to be two x minus one over x squared plus one integrated with respect to x. Then we would add on the integral of this function, three x over x squared plus two, integrated with respect to x. Now let's start by dealing with this one right about here. Well, if I want to integrate this, well, what I can do is I can think about what would allow me to integrate this function. Is there some way I could simplify this in any way? Well, I can actually break this down into two separate integrals. So if I'm integrating two x minus one over x squared plus one, that's gonna be the same thing as integrating two x over x squared plus one minus one over x squared plus one. This is just a general rule of denominators, because since we have a common denominator, we could split up the numerator. You could go ahead and resubtract these values, and you'd notice we'd still get the same thing we had before. But now I have something I can deal with. Because if I'm looking at integrating this piece, the two x over x squared plus one with respect to x, I can just do a variable substitution right about here. I can go ahead and say that this piece is equal to u, meaning u is x squared plus one. And the derivative of this would be two x with respect to x. Notice we have the two x d x right there, and we went ahead and set this equal to u. So what I could do is rewrite this as the integral of du over u. And I know that this just comes out to the natural log of u plus c where we went ahead and said that our natural log of u or or that our u right there is just going to be x squared plus one, so we'd have the natural log of x squared plus one plus the constant c. So that's what happens when we deal with this piece of the integral. We get the natural log of x squared plus one. And I'm not gonna worry about the plus c, because I'm just gonna add one big constant at the end here when we integrate both of these, because now I need to integrate this piece. But this piece is a little bit more straightforward to integrate because notice this actually follows a very similar form. This is actually the integral will come out to the inverse tangent. This is the exact same form that we've seen previously. So you can just say it's gonna be minus the inverse tangent of x. That's what happened when we have one over x squared plus one that we're integrating. It's the exact form as the inverse tangent. So that's going to be what this integral comes out to when we're just talking about this portion right about here. Now next, I need to solve for this portion of the integral. Here, we're integrating three x over x squared plus two. How would I deal with this piece? Well, if I want to integrate this function, three x over x squared plus two, what I can do is another variable substitution. Now since I already used the variable u, I'm now going to make this the variable v. So I'm gonna take my v and say that it's equal to this denominator, x squared plus two. Now if I take the derivative of this denominator, that's gonna give me two x with respect to x. Now notice here how this three is a constant, so that can come out to the front. So we'd have three times the integral of x over x squared plus two. Now we said that this x squared plus two was going to be v because that's what we set it equal to. And then we have x dx. Now notice that we said that dv is equal to two x dx. So what I can do is multiply this whole thing by a two and then divide it by a two as well, giving me that this whole piece is going to be my d v, and then we have our two be on bottom here. And I can go ahead and pull that two also out to the front. So dealing with this is going to be three halves times the integral, and then it's going to be on top, we'll have d v, and then on bottom, we'll just have v. So I know that this whole thing is just gonna come up to three halves times the natural log of the absolute value of v, where v is going to be this term right here. So we have three halves times the natural log of the absolute value of x squared plus two. And so this is what we get when we go ahead and integrate this portion three x squared three x over x squared plus two, which is this part right here. So now that I've gotten all of this figured out, I can write out the final integral right here. So the final integral is going to be this portion, l n natural log x squared plus one minus the inverse tangent of x. Then we're going to have plus three halves times the natural log of x squared plus two. So we'll have times the natural log of the absolute value of x squared plus two. And since this was the integral that we started with at the problem, we just split this into fractions right here. It's the same integral that we started with at the very beginning of this problem. We took this, and we split it into these fractions. And then we got that this was our solution right here. So this is going to be the solution to the problem, which I'll go ahead and write up here. And the final answer to the problem is going to be this result, and that is the solution and how we can solve this partial fraction decomposition strategy we needed to use to integrate this rational function. So hope you found this video helpful, and let's move on.

Use the method of partial fractions to evaluate the integral. ∫ 5 x 3 − x 2 + 7 x − 2 ( x 2 + 1 ) ( x 2 + 2 ) d x \int_{}^{}\frac{5x^3-x^2+7x-2}{(x^2+1)(x^2+2)}dx

ln ⁡ ∣ x 2 + 1 ∣ − t a n − 1 ( x ) + 3 2 ln ⁡ ∣ x 2 + 2 ∣ + C \ln\left|x^2+1\right|-tan^{-1}\left(x\right)+\frac32\ln\left|x^2+2\right|+C

5 ln ⁡ ∣ x 2 + 1 ∣ − t a n − 1 ( x ) + 3 ln ⁡ ∣ x 2 + 2 ∣ + C 5\ln\left|x^2+1\right|-tan^{-1}\left(x\right)+3\ln\left|x^2+2\right|+C

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

Struggling with Calculus?

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Multiple Choice

Use the method of partial fractions to evaluate the integral.
$\int_{}^{} \frac{5 x^{3} - x^{2} + 7 x - 2}{(x^{2} + 1) (x^{2} + 2)} d x$

$\ln ∣ x^{2} + 1 ∣ - \ln ∣ x^{2} + 1 ∣ + \frac{3}{2} \ln ∣ x^{2} + 2 ∣ + C$

$\ln ∣ x^{2} + 1 ∣ - t a n^{- 1} (x) + \frac{3}{2} \ln ∣ x^{2} + 2 ∣ + C$

$5 \ln ∣ x^{2} + 1 ∣ - \ln ∣ x^{2} + 1 ∣ + 3 \ln ∣ x^{2} + 2 ∣ + C$

$5 \ln ∣ x^{2} + 1 ∣ - t a n^{- 1} (x) + 3 \ln ∣ x^{2} + 2 ∣ + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a rational function where the denominator is a product of two irreducible quadratic factors, $(x^2 + 1)(x^2 + 2)$. This suggests using the method of partial fractions to decompose the integrand.

Step 2: Set up the partial fraction decomposition. Assume $\frac{5x^3 - x^2 + 7x - 2}{(x^2 + 1)(x^2 + 2)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 2}$, where $A, B, C,$ and $D$ are constants to be determined.

Step 3: Multiply through by the denominator $(x^2 + 1)(x^2 + 2)$ to eliminate the fractions. This gives $5x^3 - x^2 + 7x - 2 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)$. Expand and collect like terms to form a polynomial equation.

Step 4: Equate the coefficients of like powers of $x$ on both sides of the equation to form a system of linear equations. Solve this system to find the values of $A, B, C,$ and $D$.

Step 5: Substitute the values of $A, B, C,$ and $D$ back into the partial fraction decomposition. Integrate each term separately. For terms like $\frac{Ax}{x^2 + 1}$, use substitution, and for terms like $\frac{B}{x^2 + 1}$, recognize the derivative of $\tan^{-1}(x)$. Combine the results to write the final solution, including the constant of integration $C$.

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