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

Question

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

Aurora Alvarado · Accepted Answer

I have solved this way. See the answer below:

Axel Alvarez · Answer

undefined 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 $\int e^{x^3} \, dx$.