whats the integral of e^(3sqrt(x))?
i think i'm supposed to use substitution but i don't know what to substirute, i tried u= x^(1/3) but it didn't work out, please help!
To solve the integral of e^(3sqrt(x)), you're on the right track by considering substitution. Let's go through the process step by step.
1. Start by letting u = sqrt(x). This substitution aims to eliminate the square root from the equation.
2. To find du (the differential of u), differentiate both sides of the equation u = sqrt(x) with respect to x: du/dx = 1/(2*sqrt(x)).
3. Rearrange the equation to solve for dx: dx = 2*u*du.
4. Now substitute these new expressions for u and dx back into the integral:
∫ e^(3sqrt(x)) dx = ∫ e^(3u) * 2u * du.
5. Simplify the integral: ∫ 2u * e^(3u) du.
Now, this integration can be done using integration by parts. Integration by parts is a technique used to split the integrand into two functions and apply a specific rule. The rule states that if you integrate the product of two functions, you can obtain an answer by integrating one of the functions and differentiating the other.
6. According to the integration by parts rule, let u = u, and dv = 2u * e^(3u) du.
Then, du = du, and v = (1/3) * e^(3u) (with integration).
7. Apply the integration by parts formula: ∫ u * dv = u * v - ∫ v * du.
This gives us:
∫ 2u * e^(3u) du = u * [(1/3) * e^(3u)] - ∫ [(1/3) * e^(3u)] * du.
8. Continue simplifying:
∫ [(1/3) * e^(3u)] * du can be solved using a simple substitution.
Let w = 3u, then dw = 3du, and du = dw/3.
9. Substitute u back in terms of x: w = 3sqrt(x), dw = (3/2) * dx, and dx = (2/3) * dw.
10. Rewrite the integral using w:
∫ [(1/3) * e^(w)] * (2/3) * dw = (2/9) ∫ e^w dw.
11. Now integrate with respect to w:
(2/9) ∫ e^w dw = (2/9) * e^w + C, where C is the constant of integration.
12. Replace w with 3sqrt(x):
(2/9) * e^(3sqrt(x)) + C.
Hence, the integral of e^(3sqrt(x)) is (2/9) * e^(3sqrt(x)) + C, where C is the constant of integration.
Did you try letting u = 3sqrt x?
x = (1/9)u^2
dx = (2/9) u du
Now the integral becomes the integral of
(2/9) u e^u du.
You will still need another step. I suggest using integration by parts.
I get (2/9)[ u e^u - e^u] + C
= (2/9)[3sqrtx e^(3sqrtx) - e^(3sqrtx)] + C
but check my numbers. I make mistakes all the time.