Evaluate the integrals in Exercises 87–96.
95. ∫₂⁴ x^(2x...

Question

Evaluate the integrals in Exercises 87–96.

95. ∫₂⁴ x^(2x) (1 + ln x) dx

Accepted Answer

Welcome back everyone. Evaluate the d integral from 1 to 3 of x to the power of 3 x multiplied by 3 plus 3 ln of x multiplied by dx. For this problem we're going to use the u substitution. Let's suppose that u is equal to x, raise to the power of 3 x, and let's remember that our next step is to find du u divided by dx. That's the derivative of u with respect to x, and it is equal to the derivative of x to the power of 3 x. Notice that we have a form of x to the power of a function of x, right? So to find the derivative we are going to use logarithmic differentiation. Let's simply suppose that y is equal to x to the power of 3. X, right? So what we're going to do is simply take ln of both sides and show that ln of y equals ln of x, the power of 3 x. Using logarithmic differentiation, we can use the power rule, right, and we can bring down the exponent. Therefore we can now show that ln of y equals 3 x multiplied by ln of x, and from here we can take the derivative of both sides so that the derivative of ln of y is equal to the derivative of 3 x ln of x. On the left side we get 1 divided by y multiplied by y because y is a function of x, so we're applying the chain rule. On the right hand side we want to apply the product rule. So that would be 3 x derivative multiplied by ln of x plus 3 x multiplied by ln x derivative. Simplifying, we get y divided by y equals the derivative of 3 x is 3, so we get 3 ln x plus 3 x multiplied by 1 divided by x, which is the derivative of the natural log of x. And because x divided by x is 1, we simply get plus 3 multiplied by 1, which is 3. Now from here. We can notice that we have that term within our integral 3 + 3 lm of x, right? So first of all, we can show that y, which is basically u as well, or dU divided by dx. It is going to be equal to. Why? Multiplied by 3 ln of x plus 3. And we know what y is. It is x to the power of 3 x. So now d u divided by dx can be written as. X is the power of 3 x multiplied by 3 ln x plus 3, and this is our integrand right multiplying both sides by d x we get d u equals x to the power of 3 x multiplied by 3 ln x plus 3 d x. Well then, so our integrant is simply d u. Our next step is to find the new limits of integration. The lower 1 u1 is going to be equal to. X to the power of 3 x when x is 1, so that's 1 to the power of 3 multiplied by 1, that's 1 cubed, which is 1, and the upper limit u2 is going to be. 3 to the power of 3 multiplied by. 3 And this is equal to 3 to the power of 9. So now we are going to integrate the EU. From 1 up to 3 to the power of 9. Let's write that more clearly. Our integral becomes integral from 1 up to 3 to the power of 9. Of the EU And the integral of the you is simply you. And now we want to evaluate you from 1 up to 3 to the power of 9. Applying the fundamental theorem of calculus, part 2, we get 3 to the power of 9 minus 1. There's no necessity to calculate the exact value. If we want, we can do that, but this is a better form because it's more readable and Otherwise we get a large number. So if we leave it in this form, 3 is the power of 9 minus 1, it is perfectly acceptable. Thank you for watching.

Evaluate the integrals in Exercises 87–96.95. ∫₂⁴ x^(2x) (1 + ln x) dx

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Evaluate the integrals in Exercises 87–96.
95. ∫₂⁴ x^(2x) (1 + ln x) dx