\displaystyle[br]\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}
\displaystyle [br]\begin{equation*} I = \int_0^{\pi/4} \log \frac{2}{1 + \tan \theta} \, \mathrm{d}\theta = \frac{\pi}{4}\ln 2 - I \iff I = \frac{\pi}{8}\ln 2\end{equation*}
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\displaystyle [br]\begin{equation*} \int \frac{x + 1 - 1}{1+x} \, \mathrm{d}x = \int 1 - \frac{1}{x+1} \, \mathrm{d}x = x - \ln |x+1| + \mathcal{C}\end{equation*}
\displaystyle [br]\begin{equation*}\log f(x) = p \log x + q \log (x+1) + r \log (x+2) \Rightarrow f'(x) = f(x)\left(\frac{p}{x} + \frac{q}{x+1} + \frac{r}{x+2}\right)\end{equation*}
\displaystyle[br]\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}
\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3y^2}{(1+x^2)^{5/2}}\end{equation*}
\displaystyle [br]\begin{equation*} \frac{x^3 +x - x}{(1+x^2)^{5/2}} = \frac{x(1+x^2) - x}{(1+x^2)^{5/2}} = \frac{x}{(1+x^2)^{3/2}} - \frac{x}{(1+x^2)^{5/2}} \end{equation*}
\displaystyle [br]\begin{equation*}\log f(x) = p \log x + q \log (x+1) + r \log (x+2) \Rightarrow f'(x) = f(x)\left(\frac{p}{x} + \frac{q}{x+1} + \frac{r}{x+2}\right)\end{equation*}
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