AP DSC School Assistant Math Model Test 1

Concept:This integral can be solved by rewriting the integrand and applying integration by parts. The key is to notice that the derivative of $e^{x+\frac{1}{x}}$ is $\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}$, which appears when we split the given expression.Explanation:Step 1: Rewrite the integrand. We have $\int \left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}\,dx$. Notice that $1+x-\frac{1}{x} = 1 + x\left(1-\frac{1}{x^2}\right)$. Hence the integral becomes: $\int e^{x+\frac{1}{x}}\,dx + \int x\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}\,dx$. Step 2: Integrate the second term using integration by parts. Let $u = x$ and $dv = \left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}\,dx$. Then $du = dx$ and $v = e^{x+\frac{1}{x}}$. So $\int x\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}\,dx = x e^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}\,dx$. Step 3: Combine the results. Now the original integral becomes: $\int e^{x+\frac{1}{x}}\,dx + \left(x e^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}}\,dx\right) = x e^{x+\frac{1}{x}} + C$. Thus the integral evaluates to $x e^{x+\frac{1}{x}} + C$.Answer:Option B: $x e^{x+\frac{1}{x}} + c$