Indefinite integral of e^(2x) * sin(3x + 1) dx

To find the indefinite integral of the given function, we can use integration by parts. The formula for integration by parts is:

∫ u * v dx = u * ∫ v dx - ∫ (u' * ∫ v dx) dx

where u is the first function, v is the second function, u' is the derivative of u, and ∫ v dx is the integral of v with respect to x.

Step 1: Identify u and dv
In this case, we can choose u as e^(2x) and dv as sin(3x + 1) dx.

Step 2: Find du and v
To find du, we take the derivative of u with respect to x:
du/dx = d/dx(e^(2x)) = 2e^(2x)
So, du = 2e^(2x) dx

To find v, we integrate dv with respect to x:
v = ∫ sin(3x + 1) dx

Step 3: Evaluate v
To evaluate v, we need to solve the integral of sin(3x + 1) dx. This integral can be solved using the substitution method.

Let u = 3x + 1
Then, du = 3 dx

The integral becomes:
∫ sin(u) (1/3) du = (1/3) ∫ sin(u) du = -(1/3) cos(u) + C

Substituting back u = 3x + 1, we get:
v = -(1/3) cos(3x + 1) + C

Step 4: Apply the integration by parts formula
Using the integration by parts formula, we have:
∫ e^(2x) * sin(3x + 1) dx = u * v - ∫ (u' * v) dx

Substituting the values, we get:
∫ e^(2x) * sin(3x + 1) dx = e^(2x) * (-(1/3) cos(3x + 1)) - ∫ (2e^(2x) * (-(1/3) cos(3x + 1))) dx

Simplifying further, we have:
∫ e^(2x) * sin(3x + 1) dx = -(1/3) e^(2x) cos(3x + 1) + (2/3) ∫ e^(2x) cos(3x + 1) dx

Now, we have a new integral to solve, but we can use integration by parts again to find it.

Step 5: Repeat integration by parts
Let u = e^(2x) and dv = cos(3x + 1) dx

Find du and v using the same procedure as before.

Step 6: Evaluate the new integral
Using the substitution method, solve the integral of cos(3x + 1) dx and find v.

Step 7: Apply the integration by parts formula again
Using the integration by parts formula, apply it to the