Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• Recognizing derivatives within integrals to simplify.
• Careful algebraic manipulation to arrive at the correct form.

Step-by-Step Solution

Step 1: Rewrite the integral

We start by rewriting the given integral $I$: $I = \int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx}$ $I = \int {{e^{x + {1 \over x}}}dx} + \int {\left( {x - {1 \over x}} \right){e^{x + {1 \over x}}}dx}$ This splits the integral into two parts, which will be easier to handle separately.

Step 2: Apply Integration by Parts to the first integral

We apply integration by parts to the first integral, $\int {{e^{x + {1 \over x}}}dx}$. Let $u = x$ and $dv = \frac{1}{x}e^{x + \frac{1}{x}}dx$. Then $du = dx$ and $v = \int \frac{1}{x} e^{x + \frac{1}{x}} dx$. However, this does not simplify things, so let's try the following:

Let $u = x$ and $dv = e^{x + \frac{1}{x}} dx$. Then $du = dx$, and $v$ is not something we can easily find. Instead, we will try integration by parts on $\int e^{x+\frac{1}{x}}dx$ by letting $u=1$ and $dv=e^{x+\frac{1}{x}}dx$. This also doesn't seem to lead anywhere useful.

Instead, let's try integration by parts on the second integral, $\int (x - \frac{1}{x}) e^{x + \frac{1}{x}} dx$.

Step 3: Apply Integration by Parts to a different part of the original integral

Let's focus on $\int e^{x + \frac{1}{x}} dx$ and integrate by parts. Let $u = x$ and $dv = \frac{1}{x}e^{x + \frac{1}{x}} dx$. Then $du = dx$ and $\int dv$ is not obvious, therefore, we will not attempt this.

Instead, let's try integration by parts on $\int (x - \frac{1}{x})e^{x+\frac{1}{x}}dx$. Let $u = x$ and $dv = (1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx$. Note that $\frac{d}{dx}(x + \frac{1}{x}) = 1 - \frac{1}{x^2}$.

Let us integrate the original integral by parts. $I = \int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx = \int e^{x+\frac{1}{x}}dx + \int (x-\frac{1}{x})e^{x+\frac{1}{x}}dx$. Let $u = x$, and $dv = (1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx$. Then $du=dx$, and $v = \int (1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx = e^{x+\frac{1}{x}}$. So $\int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx = xe^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}dx$. Thus $\int (x-\frac{1}{x})e^{x+\frac{1}{x}}dx = \int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx = xe^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}dx$. Substituting this into the original equation: $I = \int e^{x+\frac{1}{x}}dx + xe^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}dx = xe^{x+\frac{1}{x}} + C$

Step 4: Relating the answer to the options The correct answer is $I = xe^{x+\frac{1}{x}} + C$. Let $\psi(x^3)$ be some function. Looking at the options, we need to somehow get $xe^{x+\frac{1}{x}}$ into the form $\frac{1}{3}[x^3 \psi(x^3) - \int x^2\psi(x^3)dx]$. If we let $\psi(x^3) = e^{x+\frac{1}{x}}$, then this is clearly not a function of $x^3$.

Let $\psi(x^3) = \frac{1}{x}e^{x+\frac{1}{x}}$. Then $\frac{1}{3}[x^3 \frac{1}{x}e^{x+\frac{1}{x}} - \int x^2 \frac{1}{x}e^{x+\frac{1}{x}}dx] = \frac{1}{3}[x^2e^{x+\frac{1}{x}} - \int xe^{x+\frac{1}{x}}dx]$. This is also incorrect.

Let's try to work backwards from the correct answer. We have $xe^{x+\frac{1}{x}} = \frac{1}{3}[3xe^{x+\frac{1}{x}}]$. Let's assume that $3xe^{x+\frac{1}{x}} = x^3 \psi(x^3) - \int x^2 \psi(x^3) dx$. Then we need to find $\psi(x^3)$ such that $3xe^{x+\frac{1}{x}} = x^3 \psi(x^3) - \int x^2 \psi(x^3) dx$. Differentiating both sides wrt x: $3e^{x+\frac{1}{x}} + 3x(1-\frac{1}{x^2})e^{x+\frac{1}{x}} = 3x^2 \psi(x^3) - x^2 \psi(x^3)$. $3e^{x+\frac{1}{x}} + 3(x-\frac{1}{x})e^{x+\frac{1}{x}} = 3x^2 \psi'(x^3) - x^2 \psi(x^3)$. $[3+3x-\frac{3}{x}]e^{x+\frac{1}{x}} = 3x^2 \psi'(x^3) - x^2 \psi(x^3)$. This doesn't seem to lead to a solution.

Let $\psi(x^3) = \frac{3}{x^2}e^{x+\frac{1}{x}}$. Then $\frac{1}{3}[x^3 \frac{3}{x^2}e^{x+\frac{1}{x}} - \int x^2 \frac{3}{x^2}e^{x+\frac{1}{x}} dx] = \frac{1}{3}[3xe^{x+\frac{1}{x}} - \int 3e^{x+\frac{1}{x}} dx] = xe^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}dx$. This is not the correct answer.

Let $\psi(x^3) = \frac{3}{x^2}e^{x+\frac{1}{x}}$. This is not a function of $x^3$. So we cannot find this.

The correct answer must be $xe^{x+\frac{1}{x}}+C$.

Common Mistakes & Tips

• Choosing the correct 'u' and 'dv' in integration by parts is crucial. Sometimes, you need to try different choices.
• Look for ways to simplify the integral by recognizing derivatives within the integrand.
• Working backward from the given answer can be a useful strategy in some cases.

Summary

We started by splitting the integral into two parts. Then, we tried integration by parts, and after some attempts, we found that integrating $\int (x - \frac{1}{x})e^{x + \frac{1}{x}} dx$ by parts was a useful approach. We let $u = x$ and $dv = (1 - \frac{1}{x^2}) e^{x + \frac{1}{x}} dx$. This led to the simplification of the integral and gave us the answer $xe^{x + \frac{1}{x}} + C$. This could be written as $\frac{1}{3}[3xe^{x+\frac{1}{x}}]$. If we let $\psi(x^3) = \frac{3}{x^2}e^{x+\frac{1}{x}}$, then the answer is $\frac{1}{3} [x^3\frac{3}{x^2}e^{x+\frac{1}{x}} - \int 3e^{x+\frac{1}{x}}dx] = \frac{1}{3} [3xe^{x+\frac{1}{x}} - \int 3e^{x+\frac{1}{x}}dx]$. This does not match any of the options. Therefore, there may be an error in the question.

Final Answer

The final answer is \boxed{xe^{x + \frac{1}{x}} + C}. However, the given correct answer is option (A).