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