Question Statement

Evaluate the integral:

$\int x^3{e}^{x^2}dx$

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Background and Explanation

This problem requires combining integration by substitution (to handle the composite function $e^{x^2}$) with integration by parts (to manage the product of polynomial and exponential terms). The key insight is recognizing that $x^3=x^2\cdot x$, allowing us to use $x^2$ as our substitution variable while the remaining factor $x$ becomes part of the differential.

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Solution

We begin by denoting the integral as $I$:

$I=\int x^3{e}^{x^2}dx$

Step 1: Rewrite the integrand

To prepare for substitution, we factor $x^3$ as $x^2\cdot x$. This separates the term that will become our substitution variable ($x^2$) from the term that will become part of the differential ($xdx$):

$I=\int x^2\cdot x\cdot e^{x^2}dx$

Or equivalently:

$I=\int x^2{e}^{x^2}\cdot xdx$

Step 2: Apply substitution

Let us make the substitution: $x^2=t$

Taking differentials of both sides: $2xdx=dt$

Solving for $xdx$: $xdx=\frac{dt}{2}$

Step 3: Transform the integral

Substituting $t=x^2$ and $xdx=\frac{dt}{2}$ into equation (1):

$I=\int t\cdot e^t\cdot \frac{dt}{2}$

Pulling the constant outside: $I=\frac{1}{2}\int t\cdot e^{t}dt$

Step 4: Apply integration by parts

We now use integration by parts on $\int t{e}^{t}dt$. Recall the formula: $\int udv=uv-\int vdu$

Choose:

• $u=t$ (so that $\frac{du}{dt}=1$)
• $dv=e^{t}dt$ (so that $v=e^t$)

Applying the formula:

$I=\frac{1}{2}\left[t\cdot \int e^{t}dt-\int \left(\int e^{t}dt\right)\frac{d}{dt}(t)dt\right]$

Evaluating the inner integrals:

$I=\frac{1}{2}\left[t\cdot e^t-\int e^t\cdot 1dt\right]$

$I=\frac{1}{2}\left[t{e}^t-\int e^{t}dt\right]$

$I=\frac{1}{2}\left[t{e}^t-e^t\right]+c$

Step 5: Factor and back-substitute

Factor out $e^t$:

$I=\frac{e^t}{2}(t-1)+c$

Now substitute back $t=x^2$:

$I=\frac{e^{x^2}}{2}\left(x^2-1\right)+c$

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Key Formulas or Methods Used

• Substitution Rule: $\int f(g(x))\cdot g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$
• Differential of Power: $d(x^n)=n{x}^{n-1}dx$ (used to get $xdx=\frac{dt}{2}$)
• Integration by Parts: $\int udv=uv-\int vdu$
• Basic Exponential Integral: $\int e^{t}dt=e^t+c$

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Summary of Steps

1. Factor the integrand: Rewrite $x^3$ as $x^2\cdot x$ to separate the substitution term from the differential term
2. Substitute: Let $t=x^2$, which gives $xdx=\frac{dt}{2}$
3. Simplify: Transform the integral to $\frac{1}{2}\int t{e}^{t}dt$
4. Integrate by parts: Apply $\int udv=uv-\int vdu$ with $u=t$ and $dv=e^{t}dt$
5. Back-substitute: Replace $t$ with $x^2$ to obtain the final answer $\frac{e^{x^2}}{2}(x^2-1)+c$