Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 48

Prove that $\displaystyle \frac{d}{dx} \left( \frac{1}{2} \tan^{-1} x + \frac{1}{4} \ln \frac{(x + 1)^2}{x^2 + 1} \right) = \frac{1}{(1 + x)(1 + x^2)}$

Solving for the left-hand side of the equation


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} \left( \frac{1}{2} \tan^{-1} x + \frac{1}{4} \ln \frac{(x + 1)^2}{x^2 + 1} \right) =& \frac{1}{2} \frac{d}{dx} (\tan^{-1} x) + \frac{1}{4} \ln (x + 1)^2 - \frac{1}{4} \ln (x^2 + 1)
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=& \frac{1}{2} \frac{d}{dx} (\tan^{-1} x) + \frac{2}{4} \frac{d}{dx} [\ln (x + 1)] - \frac{1}{4} \frac{d}{dx} [\ln (x^2 + 1)]
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=& \frac{1}{2} \cdot \frac{1}{ 1 +x^2} + \frac{1}{2} \cdot \frac{1}{x + 1} \frac{d}{dx} (x + 1) - \frac{1}{4} \cdot \frac{1}{x^2 + 1} \frac{d}{dx} (x^2 + 1)
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=& \frac{1}{2(1 + x^2)} + \frac{1}{2(x + 1)} - \frac{1}{4(x^2 + 1)} \cdot 2x
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=& \frac{1}{2 (1 + x^2)} + \frac{1}{2(x + 1)} - \frac{2x}{4 (x^2 + 1)}
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=& \frac{1}{2 (1 + x^2)} + \frac{1}{2(x + 1)} - \frac{x }{2 (x^2 + 1)}
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=& \frac{1}{2} \left( \frac{1}{1 + x^2} + \frac{1}{x + 1} - \frac{x}{x^2 + 1} \right)
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=& \frac{1}{2} \left( \frac{1 - x}{x^2 + 1} + \frac{1}{x + 1} \right)
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=& \frac{1}{2} \left[ \frac{(1 - x)(x + 1) + x^2 + 1}{(x^2 + 1)(x + 1)} \right]
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=& \frac{1}{2} \left[ \frac{\cancel{x} + 1 - \cancel{x^2} - \cancel{x} + \cancel{x^2} + 1}{(x^2 + 1) (x + 1)} \right]
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=& \frac{1}{\cancel{2}} \left[ \frac{\cancel{2}}{(x^2 + 1) (x + 1)} \right]
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=& \frac{1}{(x^2 + 1) (x + 1)}
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& \text{or}
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=& \frac{1}{(1 + x) (1 + x^2)}




\end{aligned}
\end{equation}
$