Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫x e^(3x) dx

Accepted Answer

Welcome back everyone. Evaluate the integral x e to the power of 2 x dx using integration by parts. For this problem, let's remember the formula. The integral of UDV is equal to UV minus the integral of VDU. What we're going to do is set u equals x, and DV is going to represent the remaining part, which is e to the power of 2 x d x. From here we can show that DU is equal to dx because U is equal to x and V is basically the integral of e to the power of 2 x dx, right? That would be 1/2 e to the power of 2 x. We do not have to introduce a constant of integration here, and we are ready to apply the formula. We're going to show that the integral of x e to the power of 2 x d x is going to be equal to u multiplied by v. So that would be 1/2 x. Multiplied by e to the power of 2 x minus the integral of v d u. So that would be the integral of 1/2, we can factor out 1/2 e to the power of 2 x, and we know that duu is d x. Notice again we're getting the same integral and we know its value, so the final answer is going to be 1/2 x e to the power of 2 x minus 1/22 multiplied by 1/22 e to the power of 2x. That's the value of the final integral. And we're going to add the constant of integration c. Finally, let's multiply and simplify. We are going to get x divided by 2 e to the power of 2 x minus 1/4 e to the power of 2 x plus a constant of integration c, and that's our final answer for this problem. Thank you for watching.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫x e^(3x) dx

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫x e^(3x) dx