find-the-indefinite-integrals-intleftt25-t1right-d-t

Question

Find the indefinite integrals.
$$\int\left(t^{2}+5 t+1\right) d t$$

Linh Vu · Accepted Answer

it was to find a indefinite integral armed with a thesis Krampus five day plus one. Did they record down the indefinite integral on the power form. And we could get a table and one on endless one. Plus a constant C is in this formula. And then we get for the first time again included at the poetry over three for the second term, the five we stayed the same. And do you hear the repairman? Wonderful. We have a table too long with you. And for the last one, the constant rich image touched the tea next to it s a constancy here.

Gregory Higby · Answer

right. The theme in this problem is that if you are asked to take a to the T. Power D. T. So the location of your expense important, uh The anti griddle is one divided by the natural log of T. Um times A. To the T. Power. And it&#x27;s just undoing the chamber and you do need to remember plus your constant. So the key in this problem then is to get your problem to look like that uh to the T. One plus five to the T. Power DT. So what we need to do is because there&#x27;s a multiplication right here, we need to distribute that in there before trying anything. And a rule here is that because we have to the T. Power. So there&#x27;s T. Number of twos and there&#x27;s T. Number of fives. That should be the same as two times 52 times five tee times. So it&#x27;s the same thing as 10 to the T. Power. I hope my explanation made sense why that&#x27;s true because now we can just use this rule up here. Uh put a boxer on that, that&#x27;s what we&#x27;re using. Um And it&#x27;s going to be won over natural log of two times, two to the T. Power, plus one over natural log of 10 times 10 to the T. Power. And don&#x27;t forget about plus C. You can verify that this is correct because the derivative of this will get us back to here, which we already know is equal to that, so we have the correct answer.

Robert Daugherty · Answer

Section 67 Problem number one Dealing with a lot of problems here to do with the integration by parts. So our formula for the integration by parts is the integral of you. Devi is equal to you ve minus the integral VD you and this just evolves out of the product rule. So what you want to do is pick something for you and pick something for a DV. And if you could do that So I&#x27;m looking at this interval. Make a substitution for you and one for TV. What you want for you is something that when you differentiate, it becomes simpler. What you want for Devi is something that you can integrate, uh, rather easily. So when you see a polynomial like T, that is a logical choice for you. So if you is equal to tea, then do you is equal to d T? Then that leaves Devi is equal to e to the five t d t. Then V would be the integration of that which would just be 1/5 yay to the five t. So this integral which is in the form you devi now can be written as you ve so that&#x27;s gonna be t over five. He to the five T minus the integral of the D u. So that is 1/5 e to the five t d t. And so this turns into t over five. You to the five. T minus 1 25th e to the five. See, plus a constant of integration. You can simplify this just by factoring out that, um, 1/5. So you&#x27;re gonna have one fifth, he to the five t, and that leaves you with one minus. Um, excuse me. It was t minus 1/5 okay?

Amy Jiang · Answer

Well, given the interrupt jobs won t buy times fire to Artie. You have to be you, g That&#x27;s what you equal to She&#x27;s to parachute And you use your patootie when half used for equal gente t g to have t d t is equal to 1/2 You we have you All right, All right, so you is equal to one squared, which is one musical to zero squared as you could tell sodas, then points do not change. Okay, now, evaluating this, we get five to their view on half. Oh, and, uh, five evaluated at one into. Okay, so we get I five over to Owen of five minus 5 to 0 over to a one of five. Okay. Said excuse me. This is actually a once a five month is one is four over to which used me to over, ooo in five exhibition.

Vipender Yadav · Answer

heavy of evaluate the integral due to the power Fife deride by under route t Square place to not ditty. Now here I&#x27;m substituting under route the square blessed to as way. So if our division both aside will get one divide by two under group D squared plus two on Harry Get to be the 12 will cancel out on is equal to the way So we&#x27;re left with G divide by under route D squared less too is equals to delay. No, we could write this in Dragila&#x27;s d squared hold square and the emptying out state. So, Doctor ah t d y under route thescore place to observe carefully. This is nothing but do I system now? We&#x27;re to calculate T Square. Initially, we assume it Underpay square plus two vehicles to buy From here we can say that if I squaring waterside you get t Square plus two equals to white square from her P square would be equals two y squared minus two. Now we conceptual level is here on the integral become Why square minus two Hold square and there is nothing What do you weigh now again? You will read this expression this enough. The function and they tickle becomes Why did a powerful bless four minus four way dot de i And this will be good shoe. Why did the bottom five divided by five plus four way minus hair? It would be wise Square should be there on Dhere, We get vie it forwarded about three divided by a tree. Bless. See where Caesar contravene relation that we have include because the integral is indefinite integral are simply there to plug really off Why? So it will be quilts to why we assumed as under root two square plus two to the power Fife Divide by five plus four Wife is a T square plus two minus four. Why is the school blessed to cube? Divide by three Blessed See So this is the answer for the intruder Due to the far Fife divide by underfoot T square Blessed to not ditty I hope you have the problem. Thank you

Linh Vu · Answer

entirely indefinite integral inventory over three months, 230 square victory, a record and, uh, barrel for the tibia and d d will go to the Tiber and less one off and risk one plus a constancy. And in energy of lettuce formula, first of community and everything. Judah power form. So have a three d bermanis Go on, minus do de Bell ministers. You de de how, however, follows first time here we cannot not this formula, because we will have that. If anybody zero in the denominator we have another formula will be won. Number T It is a coach, Adam, after a day, therefore we give the 17 contest. Um, now we three times one of a day and now this one We have a constant sentenced them times and then of the absolute t minus you stayed the same is only a kind of formula. Handing began that t power minus watch Children&#x27;s one would be minus one over minus one. Classic constancy we consume If I together mind you again. That three a round of tea minus do less to poverty Press. See here

Problem

Find the indefinite integrals. $$\int 5 e^{z} d z…

Problem 56 Easy Difficulty

Find the indefinite integrals.
$$\int\left(t^{2}+5 t+1\right) d t$$

Answer

$\frac{1}{3} t^{3}+\frac{5}{2} t^{2}+t+C$

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Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. Gleason

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Anna Marie V.

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Kayleah T.

Caleb E.

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$\frac{1}{3} t^{3}+\frac{5}{2} t^{2}+t+C$

Applied Calculus

Anna Marie V.

Heather Z.

Kayleah T.

Caleb E.

Integration Review - Intro

Definite vs Indefinite

Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. GleasonApplied Calculus

Anna Marie V.

Heather Z.

Kayleah T.

Caleb E.

Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. Gleason

Applied Calculus