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U-substitution is a method used to simplify integrals, especially when dealing with composite functions. The key idea is to transform a complicated integral into a simpler one by making a substitution. You typically substitute the inner function of a composite function with a new variable. This can transform the integral into a more recognizable form that is easier to solve.

In the given problem, we have: \(\text{Given the integral} \ \ \int \frac{x^{2}}{e^{x^{3}}} dx \).

We choose the substitution \(u = x^3\). Why? Because the derivative of \(u\), which is \(du\), is directly related to the \(x^2 dx\) term present in the integral. This substitution will help us transform the integral into a simpler form.

Exponential functions, especially those involving the natural exponent (Euler's number \(e\)), appear frequently in integrals. The function \(e^x\) has a unique property: its derivative and its integral are both \(e^x\).

When you have an exponential function in the denominator, like \(e^{x^3}\) in the given problem, it often simplifies well with substitution. In this problem, after performing the u-substitution \(u = x^3\), the integral transforms to involve \(e^{-u}\), which is simpler to integrate.

It turns out that: \(\text{The integral of } e^{-u} = -e^{-u} + C \).

This reduces the complexity of the problem, making it easier to find a solution.

Integration is one of the core operations in calculus, and various techniques like u-substitution, integration by parts, partial fraction decomposition, and more are available to tackle different types of integrals. For this particular problem, the appropriate method is u-substitution.

Here’s a quick recap of how we solved it:

• First, we identified that the integral could be simplified with a u-substitution.
• We set \(u = x^3\), and thus \(du = 3x^2 dx\), or \(\frac{1}{3}du = x^2 dx\).
• We substituted these expressions into the integral, converting it into \(\frac{1}{3} \times \frac{1}{e^u} du\).
• This simplified to \(\frac{1}{3} \times e^{-u} du\), an easy integral.

We then integrated \(e^{-u}\), substituted back the original variable, and arrived at the final solution.

Understanding and practicing various integration techniques can significantly improve your problem-solving skills in calculus.