To solve this integral, we can use integration by parts. The integration by parts formula states:

∫u dv = uv - ∫v du

Let's assign u = x^2 and dv = e^x dx. Then, we can find du and v:

Taking the derivative of u, we get du/dx = 2x. To find v, we integrate dv = e^x dx. The integral of e^x is simply e^x, so v = e^x.

Now, we can apply the integration by parts formula:

∫x^2 e^x dx = x^2 e^x - ∫2x e^x dx

The remaining integral, ∫2x e^x dx, can be solved using integration by parts again. Assigning u = 2x and dv = e^x dx, we find du = 2 dx and v = e^x. Applying the integration by parts formula once more:

∫2x e^x dx = 2x e^x - ∫2 e^x dx

The integral ∫2 e^x dx is simply 2 e^x, so the final result is:

∫x^2 e^x dx = x^2 e^x - 2x e^x + 2 e^x + C

where C is the constant of integration.

For example, if we want to evaluate the integral from 0 to 1, we substitute the limits of integration into the result:

∫[0,1] x^2 e^x dx = (1^2 e^1 - 2(1) e^1 + 2 e^1) - (0^2 e^0 - 2(0) e^0 + 2 e^0)

Simplifying, we obtain:

∫[0,1] x^2 e^x dx = (e - 2e + 2e) - (0 - 0 + 2)

∫[0,1] x^2 e^x dx = e

Therefore, the value of the integral from 0 to 1 is e.

For references and further reading on integration techniques, you may refer to:

1. Stewart, J. (2007). Calculus: Early Transcendentals. Cengage Learning.
2. Khan Academy. (n.d.). Integration by parts. Retrieved from https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-2/v/integration-by-parts-introduction

I hope this explanation helps! Let me know if you have any further questions.