# What is #int_1^ln5 xe^(x^2)+x^2e^x+x^3+e^(x^3) dx#?

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To solve the integral (\int_1^{\ln5} xe^{x^2} + x^2e^x + x^3 + e^{x^3} , dx), we first evaluate each term of the integrand separately and then integrate each term separately using the rules of integration.

Let's denote (f(x) = xe^{x^2} + x^2e^x + x^3 + e^{x^3}).

(\int xe^{x^2} , dx) can be solved using the substitution method, letting (u = x^2) which implies (du = 2x , dx), so (dx = \frac{1}{2}du).

(\int xe^{x^2} , dx = \frac{1}{2} \int e^u , du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C).

Next, (\int x^2e^x , dx) can be solved using integration by parts. Let (u = x^2) and (dv = e^x , dx), then (du = 2x , dx) and (v = e^x).

(\int x^2e^x , dx = x^2e^x - 2\int xe^x , dx = x^2e^x - 2(e^x + C) = x^2e^x - 2e^x + C).

For (\int x^3 , dx), it's a simple power rule integration.

(\int x^3 , dx = \frac{1}{4}x^4 + C).

Lastly, (\int e^{x^3} , dx) cannot be expressed in terms of elementary functions, so it remains as (\int e^{x^3} , dx).

Now, we integrate each term from 1 to (\ln5) and sum them up:

(\frac{1}{2} \left(e^{(\ln5)^2} - e^1\right) + \left(\ln5e^{\ln5} - 2e^{\ln5} - \ln5 + 2\right) + \frac{1}{4}(\ln5)^4 - \frac{1}{4} + \int_1^{\ln5} e^{x^3} , dx).

This integral cannot be expressed in terms of elementary functions, so it remains in the form of (\int e^{x^3} , dx).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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