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Integrate the function
$\frac{2 x}{1+x^2}$
$\frac{2 x}{1+x^2}$
Solution:
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Verified Answer
Let $1+x^2=t \Rightarrow 2 x d x=d t$
$\begin{aligned}
\therefore \quad & \int \frac{2 x}{1+x^2} d x=\int \frac{d t}{t}=\log t+\mathrm{C} \\
=& \log \left(1+x^2\right)+\mathrm{C}
\end{aligned}$
$\begin{aligned}
\therefore \quad & \int \frac{2 x}{1+x^2} d x=\int \frac{d t}{t}=\log t+\mathrm{C} \\
=& \log \left(1+x^2\right)+\mathrm{C}
\end{aligned}$
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