Evaluate the integral.∫t4+t2−2t3+tdt\int_{}^{}\frac{t^4+t^2-2}{t^3+t}dt

t 2 2 − 2 ln ⁡ ∣ t ∣ + ln ⁡ ∣ t 2 + 1 ∣ + C \frac{t^2}{2}-2\ln\left|t\right|+\ln\left|t^2+1\right|+C

Evaluate the integral. ∫ t 4 + t 2 − 2 t 3 + t d t \int_{}^{}\frac{t^4+t^2-2}{t^3+t}dt

Let's give this problem a try. Here we're asked to evaluate the following integral that we see. Now I notice that the integral that we're dealing with here is t to the fourth plus t squared minus two over t cubed plus t. Now if we're integrating this whole thing, it may be tempting to start with a variable substitution, since we can see we have a kinda tricky looking rational function. If you tried a variable substitution on the denominator, you would get u is equal to t cubed plus t, and taking the derivative would give us three t squared plus one. That's not going to help us with anything that we have in the numerator, so that's not gonna work. We could try a variable substitution on the numerator as well, but doing this is not going to help us with the denominator in any way, so that's not gonna be our strategy for solving this problem. But what we can do is we can try breaking this up now into partial fractions to see if that's going to be a strategy that allows us to solve this. Now right off the bat, when it comes to the partial fraction strategy, I'm first going to take a look at the highest degree in my numerator, which I see is four t to the fourth power, and the highest degree in my denominator is three t to the third power. So because of that, this numerator has a higher degree than the denominator, so what we first want to do is perform long division before we go ahead and set up our partial fractions. So let's go ahead and do that. So So what we're gonna have is t cubed plus t that we are dividing into this whole piece in the numerator, which is t to the fourth plus t squared minus two. Now what times t cubed is going to give us this t to the fourth? Well, I think that makes sense that it would just be t, because t times t cubed is t to the fourth power, and t times t is t squared. I can go ahead and subtract these results right here, and notice if we were to subtract these, these pieces would just cancel entirely with those, so all we would be left with down here is negative two. So what that means is we can now split this into separate fractions, this is the same thing as the integral of of in this case, we got t, so we would have the integral of t, and then we would have minus this remainder two over t cubed plus t. Now keep in mind, this is actually gonna be plus negative two since since I see we have a negative two as a remainder, but I went ahead and just changed that to minus automatically since I already know that it's just gonna be minus this whole thing. So this is now the integral that we are dealing with. Now at this point, I know how to integrate just t, we can use the power rule there, so I'm not even too worried, so I'm not too worried about that at all. But how do we integrate this two over t cubed plus t? Well, thankfully, now that we've done our long division step, notice that the numerator is just a constant, and it's being divided by t cubed plus t. So we no longer have to do any long division simplification here, but what we will need to do is find some sort of way to integrate this. And so what I'm gonna do is write this fraction by itself. So if I write this two over t cubed plus t by itself, notice I can factor this denominator by pulling a t out. So we'd end up with two over t times t squared plus one. Now notice this kinda looks like a scenario where we would have to use partial fractions, and that's actually exactly what we need to do to solve this. So what I can recognize is that this breaks down into t, which is a distinct linear factor, so we'd end up with some constant a over that factor t, plus t squared plus one, which is an irreducible quadratic factor. So since it's an irreducible quadratic factor, we'd end up getting a linear equation rather than a constant, so b x plus c over, actually that would be b t plus c because we're dealing with t's this time, so b t plus c over, in this case, it would be t squared plus one. So that's what we end up getting. So now that we've gone ahead and simplified this and broken things down, and keep in mind that this whole piece that we have on the left side here, this is simply just our way of writing a factored version of this right here. So since we have this all broken down like so, now we need to go ahead and try and solve for our constants a, b, and c. Well, to do that, I'm gonna take both sides of this equation, and I'm going to multiply it by the common denominator. In this case, I can see the common denominator is t times t squared plus one. So let's multiply this out into the brackets right here. Now first off I'm going to multiply this common denominator by the two over t times t squared plus one. I can see right here that that denominator is going to cancel entirely, so all we're going to have left on the left side of the equal sign is two. Now next I'm going to move to this a term. If I multiply this by this common denominator, notice the t entirely goes away, so all we're gonna be left with right about there is a times t squared plus one. Now next we're going to move to this last piece right about here, which multiplying this by the common denominator, I can see the t squared plus one will go away entirely. So all I'm going to be left with right about here is b t plus c times t. So this is where we're at. Now, this is a case where you could either try to make some substitutions for t, or you can try to expand everything out and get a system of equations. Personally, a lot of times with these problems when doing irreducible quadratic factors, I actually like a system of equations better. I know that the variable substitution for t can sometimes be helpful, or x or whatever variable you have. I know that can sometimes be straightforward when it comes to these types of problems, but when you have these irreducible quadratic factors, a lot of times things get really complicated when you try to do substitutions. And I find it can be a little easier at the end when you just have everything written as a system. So let's go ahead and do that. Now I'm gonna distribute this a into the parentheses right here, giving us a times t squared plus a, and I'm going to distribute the t into those parentheses, giving us plus b t squared plus c t. Now at this point I can group terms together. I can see that we have an a t squared here, and a b t squared right there. So I can group those to be a plus b times t squared. Now next I'm going to look for just t's. Well, I can see we have just a t right about there, so that's going to give us this plus c t, and then I can see we just have a constant a right about there, so that's just going to be plus a. So that's what happens when we go ahead and group things together, and I want you to notice something. We have a t squared, we have a t, and then we have this constant a, but notice on the left side, we don't have a t squared or a t. We only have a constant two. So what I'm going to do is add a zero t squared plus a zero t plus two, and this is perfectly okay to do, because zero times anything is just zero, so this is the exact same argument we had before. But now I can equate these coefficients in a more easy manner. So I can say that everything in front of the t squared here is equal to everything in front of the t squared there. So we can just set up a system that says a plus b is equal to zero. Now we have one system of equations, or one equation, I should say, set up. Now next, I'm going to move to the t's. Now I can see right about here that we have c t, and since I can see that c is the coefficient in front of t, I'll set it equal to this zero right here, so we can say that c is equal to zero, and we have our second equation. Now lastly, I'm going to move to this constant a. I can see we have this constant right here, and then we have one constant right there, so I can just go ahead and say that a equals two. Now at this point, this is actually really helpful, because because notice by writing out the system of equations, we were already able to solve for the a, which is equal to two, and we were already able to solve for the c, which equals zero. The only thing we didn't have solved for was b, but if we have this equation right here, I can recognize that a is two, so we just have two plus b equals zero, and subtracting that two on both sides of the equation gives us b is equal to negative two. So now we have our a, our b, and our c. So now that we have these variables, well, I can go ahead and replace them in the equation that we had up here in these fractions that we split up. So doing this, I'll take that a, and I'll replace it with two. I'll take that b, and I'll replace it with a negative two. And then I'll take that c, and I'll just remove it completely since that c is zero. So this is gonna be negative two t, which I'm gonna write this all as just minus two t. And now we have this fraction split up. So if I were to go back up here to the case that we started with, which was this fraction right here, well, we already know the integral of t. That is done. But if I wanna go ahead and write out the rest of this fraction, well, that's going to be equal to, in this case, gonna be the integral of t, and then we split this up into that. So splitting this up, we have that this is going to be minus, and in this case, it's gonna be minus this whole piece right here, and so it's gonna be minus what we plug in right here, which is gonna be two over t minus two t over t squared plus one, all integrated with respect to t. And I just noticed that I wrote a d x here earlier. This should really be a d t. So we're integrating this all with respect to t, so that is the variable that we're focusing on. So now that we have everything in terms of t, well, what I can do is I can distribute this negative sign into the brackets right here just to make this a little bit more simple. So we're going to have t minus, and then it's gonna be two over t. And then minus this negative will become plus, and then it's going to be two t over t squared plus one. And all of this is integrated with respect to t. Now how do I do this integration? Well, to do this, I'll go ahead and move this down here since we already did that long division step, and I just need to integrate each one of these pieces separately. So first off, I'm gonna integrate the t. Integrating t using the power rule is gonna give us t squared over two. Then we're going to have minus, and then if I go ahead and integrate this, well, this is just a constant, so I can pull that out, and the integral of one over t with respect to t is just the natural log of the absolute value of t. So I get that, and then all I'm left with is this piece right here. But how do I integrate this two t over t squared plus one? Well, this one's actually surprisingly straightforward, because if you're integrating two t over t squared plus one, what you can just do is a variable substitution down there. So we can say that u is equal to t squared plus one, and the derivative would be two t with respect to t. So writing this all out, we'd have the integral of one over u, and then two two t dt is du. So we just need to integrate one over u with respect to u, which would give us the natural log of the absolute value of u plus c. And I can always see the natural log of u is just gonna be the natural log of t squared plus one plus c. So if I were to write this all out, this whole piece would come out to plus the natural log of t squared plus one. The absolute value bars, you don't wanna forget those. And then we would have plus the constant of integration c, giving us the solution to this problem. So that's how you can solve this case and this problem. Hope you found this video helpful.

Evaluate the integral. ∫ t 4 + t 2 − 2 t 3 + t d t \int_{}^{}\frac{t^4+t^2-2}{t^3+t}dt

t 2 2 − 2 ln ⁡ ∣ t ∣ + ln ⁡ ∣ t 2 + 1 ∣ + C \frac{t^2}{2}-2\ln\left|t\right|+\ln\left|t^2+1\right|+C

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

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

t 2 2 − 2 ln ⁡ ∣ t ∣ − ln ⁡ ∣ t 2 + 1 ∣ + C \frac{t^2}{2}-2\ln\left|t\right|-\ln\left|t^2+1\right|+C

12. Techniques of Integration

Partial Fractions

Calculus

12. Techniques of Integration

Partial Fractions

Struggling with Calculus?

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Multiple Choice

Evaluate the integral.
$\int_{}^{} \frac{t^{4} + t^{2} - 2}{t^{3} + t} d t$

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

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

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

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

Verified step by step guidance

Step 1: Simplify the given integral. The integral is ∫(t^4 + t^2 - 2)/(t^3 + t) dt. Start by factoring the denominator t^3 + t as t(t^2 + 1). This will help simplify the fraction.

Step 2: Perform polynomial long division if necessary. Divide the numerator (t^4 + t^2 - 2) by the denominator (t(t^2 + 1)) to separate the integral into simpler terms. This will result in a combination of polynomial terms and a proper fraction.

Step 3: Decompose the proper fraction. For the remaining fraction, use partial fraction decomposition to express it as a sum of simpler fractions. This will make it easier to integrate each term individually.

Step 4: Integrate each term. For the polynomial terms, use the power rule for integration: ∫t^n dt = t^(n+1)/(n+1). For the logarithmic terms, recognize forms like ∫1/t dt = ln|t|. For terms involving t^2 + 1, recognize that ∫1/(t^2 + 1) dt = tan^(-1)(t).

Step 5: Combine the results. Add all the integrated terms together, including the constant of integration C. Simplify the expression to match one of the given answer choices.

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