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Find the integrals.$$\int t e^{5 t} d t$$

$$\frac{1}{5} t e^{5 t}-\frac{1}{25} e^{5 t}+C$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

Antiderivatives and Applications

Section 7

Integration by Parts

Integrals

Integration

Integration Techniques

Applications of Integration

Baylor University

University of Michigan - Ann Arbor

Boston College

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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'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'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'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?

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