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$\int\left\{\frac{(\log x-1)}{\left(1+(\log x)^2\right.}\right\}^2 d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{x}{(\log x)^2+1}+C$
$\frac{x}{(\log x)^2+1}+C$
$$
\begin{aligned}
& \int \frac{(\log x-1)^2}{\left(1+(\log x)^2\right)^2} d x \\
& =\int\left[\frac{1}{\left(1+(\log x)^2\right)}-\frac{2 \log x}{\left(1+(\log x)^2\right)^2}\right] d x \\
& =\int\left[\frac{e^t}{1+t^2}-\frac{2 t e^t}{\left(1+t^2\right)^2}\right] d t \text { put } \log x=t \Rightarrow d x=e^t d t \\
& \int e^t\left[\frac{1}{1+t^2}-\frac{2 t}{\left(1+t^2\right)^2}\right] d t \\
& =\frac{e^t}{1+t^2}+c=\frac{x}{1+(\log x)^2}+c
\end{aligned}
$$
\begin{aligned}
& \int \frac{(\log x-1)^2}{\left(1+(\log x)^2\right)^2} d x \\
& =\int\left[\frac{1}{\left(1+(\log x)^2\right)}-\frac{2 \log x}{\left(1+(\log x)^2\right)^2}\right] d x \\
& =\int\left[\frac{e^t}{1+t^2}-\frac{2 t e^t}{\left(1+t^2\right)^2}\right] d t \text { put } \log x=t \Rightarrow d x=e^t d t \\
& \int e^t\left[\frac{1}{1+t^2}-\frac{2 t}{\left(1+t^2\right)^2}\right] d t \\
& =\frac{e^t}{1+t^2}+c=\frac{x}{1+(\log x)^2}+c
\end{aligned}
$$
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