Evaluate the indefinite integral.∫3tt2+7dt\int_{}^{}3t\sqrt{t^2+7}dt∫3tt2+7dt

( t 2 + 7 ) 3 2 + C \left(t^2+7\right)^{\frac32}+C

Evaluate the indefinite integral. ∫ 3 t t 2 + 7 d t \int_{}^{}3t\sqrt{t^2+7}dt ∫ 3 t t 2 + 7 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> d t

Here we're asked to evaluate the indefinite integral of three t times the square root of t squared plus seven dt. So getting started with our substitution process here, the first thing that we wanna do is choose u and find du. Now remember, u is often going to be an inside function in parentheses under a radical, etc. So here we can see that we have this t squared plus seven under a radical, which is a good clue that this would be a good choice for you. So if we make u equal to that t squared plus seven, what does that make du? Well du is the derivative of that t squared, seven times dt here since that's the variable that we're working with. So this is going to be two t dt. So moving on to our second step here rewriting our integral only in terms of this u and du, let's look at what's going on with our integral. We have this u right here and then we also have t and dt that we know are part of du, but we need this to be two t dt. So let's go ahead and rewrite this integral here, putting that t and dt together so that we can visualize this a bit better. So this is three times the square root of t squared plus seven and then times t dt. We want this to be two t dt, so we can go ahead and just stick that two in there multiplying by it. But if we multiply by a constant like we are here, we also need to multiply by that constant's reciprocal to effectively cancel it out. Now the reciprocal of two is just one half flipping that to the bottom of a fraction, so I'm just going to go ahead and stick that on the outside of my integral here because I know that I'm not going to need it inside. It's just a constant. We can just pull it out already. So looking at this now, I have a u under my radical and then I have this du. So we can rewrite this only in terms of that u and du. I'm gonna go ahead and pull that constant three out front of my integral as well, making that constant three halves. So this is three halves times the integral of u to the power of one half, just making that radical a power, times d u. So moving on to step three here, we wanna go ahead and integrate with respect to u. So we can go ahead and use the power rule here, keeping that constant out front, that's three halves, and then multiply it here by adding a one to that exponent. One half plus one is going to be three halves. So if I'm dividing here by that one half plus one, that three halves, I'm really just flipping that and multiplying here by two thirds. So this is two thirds times u to the power of three halves and then plus our constant of integration c. Now we can do some simplification here because three halves times two thirds, that's going to cancel out perfectly. Three on the top, three on the bottom, two on the top, two on the bottom. So I'm really just left with u to the power of three halves plus that constant of integration c. So moving on to our final step here, we want to replace u with our original function here of t. So in this case, remember, u was t squared plus seven. That's what we wanna replace u with here. So this is going to be t squared plus seven to the power of three halves and then plus that constant of integration c, and we're all done here. Feel Feel free to let us know if you have questions, and let's continue practicing.

Evaluate the indefinite integral. ∫ 3 t t 2 + 7 d t \int_{}^{}3t\sqrt{t^2+7}dt ∫ 3 t t 2 + 7 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> d t

( t 2 + 7 ) 3 2 + C \left(t^2+7\right)^{\frac32}+C

( t 2 + 7 ) 2 3 + C \left(t^2+7\right)^{\frac23}+C

8. Definite Integrals

Substitution for Indefinite Integrals

Business Calculus

8. Definite Integrals

Substitution for Indefinite Integrals

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}3t\sqrt{t^2+7}dt$

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

${(t^{2} + 7)}^{\frac{2}{3}} + C$

${(t^{2} + 7)}^{\frac{3}{2}} + C$

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

Verified step by step guidance

Step 1: Recognize that the integral involves a composite function, specifically the square root of (t^2 + 7). This suggests that substitution might be a useful technique to simplify the integral.

Step 2: Let u = t^2 + 7. Then, compute the derivative of u with respect to t: du/dt = 2t, or equivalently, du = 2t dt.

Step 3: Rewrite the integral in terms of u. The term 3t dt can be expressed as (3/2) du (since du = 2t dt). The square root of (t^2 + 7) becomes u^(1/2). The integral now becomes ∫(3/2) u^(1/2) du.

Step 4: Apply the power rule for integration to ∫u^(1/2) du. The power rule states that ∫u^n du = (u^(n+1))/(n+1) + C, where n ≠ -1. Here, n = 1/2, so the integral becomes (3/2) * [(u^(3/2)) / (3/2)] + C.

Step 5: Simplify the result and substitute back u = t^2 + 7 to return to the original variable. The final expression is (t^2 + 7)^(3/2) + C.

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