Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• Substitution: $\int f(g(x))g'(x) \, dx = \int f(u) \, du$, where $u = g(x)$ and $du = g'(x) \, dx$
• Form of Integral: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$

Step-by-Step Solution

Step 1: Rewrite the integrand. We begin by expanding the numerator of the integrand: \int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} = \int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} dx = \int {{{{{\left( {\log x} \right)}^2} - 2\log x + 1} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} dx

Step 2: Separate the fraction into two terms. Separate the fraction to make it easier to integrate: \int {{{{{\left( {\log x \right)}^2} - 2\log x + 1} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} dx = \int {{{1 + {{\left( {\log x \right)}^2} - 2\log x} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} dx = \int {\left[ {{{1 + {{\left( {\log x} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx = \int {\left[ {{1 \over {1 + {{\left( {\log x \right)}^2}}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} \right]} dx

Step 3: Perform a substitution. Let $t = \log x$. Then, $x = e^t$ and $dx = e^t \, dt$. Substitute these into the integral: \int {\left[ {{1 \over {1 + {{\left( {\log x \right)}^2}}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x \right)}^2}} \right)}^2}}}} \right]} dx = \int {\left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} {e^t}\,dt $= \int {{e^t}\left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt$

Step 4: Recognize the form $\int e^x [f(x) + f'(x)] dx$ We can observe that the integral is in the form $\int e^t [f(t) + f'(t)] \, dt$, where $f(t) = \frac{1}{1+t^2}$. Let's find its derivative: $f'(t) = \frac{d}{dt} \left( \frac{1}{1+t^2} \right) = \frac{-2t}{(1+t^2)^2}$ So, the integral becomes: $\int {{e^t}\left[ {{1 \over {1 + {t^2}}} + \frac{-2t}{{{\left( {1 + {t^2}} \right)}^2}}} \right]} dt = \int e^t [f(t) + f'(t)] \, dt = e^t f(t) + C = e^t \left( \frac{1}{1+t^2} \right) + C$

Step 5: Substitute back to the original variable. Substitute $t = \log x$ back into the expression: $e^t \left( \frac{1}{1+t^2} \right) + C = x \left( \frac{1}{1+(\log x)^2} \right) + C = \frac{x}{1+(\log x)^2} + C$

Step 6: Verify the given correct answer. The given correct answer is $\frac{\log x}{1 + (\log x)^2} + C$. There seems to be an error in the provided answer. Let's take derivative of $\frac{\log x}{1 + (\log x)^2}$. $\frac{d}{dx}\left(\frac{\log x}{1 + (\log x)^2}\right) = \frac{\frac{1}{x}(1 + (\log x)^2) - \log x \cdot 2 (\log x) \cdot \frac{1}{x}}{(1 + (\log x)^2)^2} = \frac{\frac{1}{x}(1 + (\log x)^2 - 2(\log x)^2)}{(1 + (\log x)^2)^2} = \frac{1}{x}\frac{1 - (\log x)^2}{(1 + (\log x)^2)^2}$ $= \frac{1}{x}\frac{1 - (\log x)^2}{(1 + (\log x)^2)^2} = \frac{1 - (\log x)^2}{x(1 + (\log x)^2)^2}$ This is not equal to the integrand, so the provided correct answer is incorrect.

Let's consider option (D): $\frac{x}{(\log x)^2 + 1} + C$. Taking its derivative: $\frac{d}{dx} \left( \frac{x}{(\log x)^2 + 1} \right) = \frac{1 \cdot ((\log x)^2 + 1) - x \cdot 2(\log x)\cdot \frac{1}{x}}{((\log x)^2 + 1)^2} = \frac{(\log x)^2 + 1 - 2\log x}{((\log x)^2 + 1)^2} = \frac{(\log x - 1)^2}{((\log x)^2 + 1)^2}$ Thus, $\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} = \frac{x}{(\log x)^2 + 1} + C$

Common Mistakes & Tips

• Remember to substitute back to the original variable after integration.
• Recognizing the form $\int e^x [f(x) + f'(x)] \, dx$ can simplify the integration process.
• Be careful when differentiating and integrating logarithmic functions.

Summary We simplified the given integral by expanding the numerator, separating the fraction, and performing a substitution. This allowed us to identify the form $\int e^x [f(x) + f'(x)] \, dx$, which simplifies the integration. After substituting back to the original variable, we obtained the result $\frac{x}{1+(\log x)^2} + C$.

Final Answer The final answer is \boxed{\frac{x}{(\log x)^2 + 1}}, which corresponds to option (D).