Use integration by parts to find the integral. Round the answer to two decimal places if necessary. (x+4)^ln x dx between 3 and 0
You sure this is right?
Even online integrators like wolfram cannot evaluate it using elementary functions
that's what I was thinking, the teacher must have made a mistake on this question ... no one I ask can help me with it.
To find the integral of (x+4)^(ln(x)), we can use integration by parts. The formula for integration by parts is:
∫ uv dx = u ∫ v dx - ∫ u' (∫ v dx) dx
Let's assign the values for u, v, u', and v' in the formula:
u = ln(x), u' = 1/x (derivative of ln(x))
v = (x+4), v' = 1 (derivative of x+4)
Now we can substitute these into the formula:
∫ (x+4)^(ln(x)) dx = ln(x) ∫ (x+4) dx - ∫ (1/x) (∫ (x+4) dx) dx
First, let's solve the simpler integral: ∫ (x+4) dx
∫ (x+4) dx = (1/2)x^2 + 4x + C
Now we can substitute this result back into the formula:
∫ (x+4)^(ln(x)) dx = ln(x) [(1/2)x^2 + 4x] - ∫ (1/x) [(1/2)x^2 + 4x] dx
Next, let's simplify this expression:
∫ (x+4)^(ln(x)) dx = (1/2)x^2 * ln(x) + 4x * ln(x) - (1/2) ∫ x ln(x) dx - 4 ∫ ln(x) dx
The remaining integrals, ∫ x ln(x) dx and ∫ ln(x) dx, can be solved using integration by parts again. However, for our specific integral (x+4)^(ln(x)), the integrals ∫ x ln(x) dx and ∫ ln(x) dx diverge as x approaches 0. Therefore, we cannot directly evaluate the integral between 0 and 3.
However, we can find the definite integral between 3 and a very small positive number, such as 0.001, where these integrals are well-behaved, and then approximate the result.
Doing so, we have:
∫ (x+4)^(ln(x)) dx = [(1/2)x^2 * ln(x) + 4x * ln(x) - (1/2) ∫ x ln(x) dx - 4 ∫ ln(x) dx] between 3 and 0.001
∫ (x+4)^(ln(x)) dx ≈ [(1/2)(0.001)^2 * ln(0.001) + 4(0.001) * ln(0.001) - (1/2) ∫ x ln(x) dx - 4 ∫ ln(x) dx] - [(1/2)(3)^2 * ln(3) + 4(3) * ln(3) - (1/2) ∫ x ln(x) dx - 4 ∫ ln(x) dx]
Simplifying further is not possible without evaluating the divergent integrals directly. Therefore, the round answer to two decimal places for the given integral cannot be determined for the given limits of 0 and 3.