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Evaluate: $\int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x$
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Explanation
$\int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x$
Let $x+1=u, x-1=u-2$
When $x=3, u=3+1=4$
\begin{aligned} &x=1, u=1+1=2 \\ &\therefore \int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x \equiv \int_{2}^{4}\left(\frac{u-2}{u^{2}}\right) \mathrm{d} u \\ &=\int_{2}^{4}\left(\frac{u}{u^{2}}-\frac{2}{u^{2}}\right) \mathrm{d} u \\ &=\int_{2}^{4}\left(\frac{1}{u}-2 u^{-2}\right) \mathrm{d} u \\ &=\left.\left[\ln u-\frac{2 u^{-1}}{-1}\right]\right|_{2} ^{4} \\ &=\left.\left[\ln u+2 u^{-1}\right]\right|_{2} ^{4} \\ &=\left[\ln u+\frac{2}{u}\right]_{2}^{4} \\ &=\left(\ln 4+\frac{2}{4}\right)-\left(\ln 2+\frac{2}{2}\right) \\ &=\ln 4-\ln 2-\frac{1}{2} \\ &=\ln \left(\frac{4}{2}\right)-\frac{1}{2} \\ &=\ln 2-\frac{1}{2} \end{aligned}
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