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arrow_back Evaluate: \(\int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x\)

by Platinum
(119,140 points)
in Mathematics
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Evaluate: \(\int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x\)

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Best answer
Explanation
\[
\int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x
\]
Let \(x+1=u, x-1=u-2\)
When \(x=3, u=3+1=4\)
\[
\begin{aligned}
&x=1, u=1+1=2 \\
&\therefore \int_{1}^{3}\left(\frac{x-1}{(x+1)^{2}}\right) \mathrm{d} x \equiv \int_{2}^{4}\left(\frac{u-2}{u^{2}}\right) \mathrm{d} u \\
&=\int_{2}^{4}\left(\frac{u}{u^{2}}-\frac{2}{u^{2}}\right) \mathrm{d} u \\
&=\int_{2}^{4}\left(\frac{1}{u}-2 u^{-2}\right) \mathrm{d} u \\
&=\left.\left[\ln u-\frac{2 u^{-1}}{-1}\right]\right|_{2} ^{4} \\
&=\left.\left[\ln u+2 u^{-1}\right]\right|_{2} ^{4} \\
&=\left[\ln u+\frac{2}{u}\right]_{2}^{4} \\
&=\left(\ln 4+\frac{2}{4}\right)-\left(\ln 2+\frac{2}{2}\right) \\
&=\ln 4-\ln 2-\frac{1}{2} \\
&=\ln \left(\frac{4}{2}\right)-\frac{1}{2} \\
&=\ln 2-\frac{1}{2}
\end{aligned}
\]
by Platinum
(119,140 points)

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