I'm having some issues with computing this indefinite integral. I don't know how to even begin. Please help! (let S be the symbol for the integral)
S (square root (x)+e^-3x) dx
I made a little bit of progress. I changed it to
S (x^1/2+e^-3x) dx
Go on:
wolframalpha dot com
When page be open in rectangle type:
integrate square root (x)+e^-3x dx
and click option =
When you see result click option:
Show steps
To compute this indefinite integral, you can use the method of integration by substitution. Here's a step-by-step guide on how to solve it:
1. Let's take a closer look at the integral:
∫(√(x) + e^(-3x)) dx
2. We can notice that the first term, √(x), can be simplified by expressing it as x^(1/2). The integral becomes:
∫(x^(1/2) + e^(-3x)) dx
3. To solve this integral, let's make a substitution. We'll substitute a new variable, u, for the inner function of the first term, u = x^(1/2). From this substitution, we have to find the differential of u, which is du = (1/2)x^(-1/2) dx.
4. We can rewrite the integral using the new variable u and its differential:
∫(2u du + e^(-3x) dx)
5. Now, we can separate the integral into two parts:
∫2u du + ∫e^(-3x) dx
6. The first part, ∫2u du, can be computed easily as:
u^2 + C1, where C1 is the constant of integration.
7. For the second part, ∫e^(-3x) dx, we can notice that it represents the integral of an exponential with respect to x. The integral of e^(kx) dx is (1/k)e^(kx) + C2, where C2 is the constant of integration. Plugging in k = -3, we get:
∫e^(-3x) dx = (-1/3)e^(-3x) + C2.
8. Now, let's put the two parts together:
∫(√x + e^(-3x)) dx = u^2 + C1 + (-1/3)e^(-3x) + C2
9. Finally, combining the constants of integration C1 and C2 into a single constant C, we obtain the final result:
∫(√x + e^(-3x)) dx = x^(1/2) - (1/3)e^(-3x) + C
Therefore, the solution to the given indefinite integral is:
∫(√(x) + e^(-3x)) dx = x^(1/2) - (1/3)e^(-3x) + C, where C is the constant of integration.