Evaluate : \(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\).
Question 1 Report
Evaluate : \(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\).
Answer Details
Rewrite the integrand by splitting \(x-1=(x+1)-2\):
\[\frac{x-1}{(x+1)^{2}}=\frac{(x+1)-2}{(x+1)^{2}}=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\]
Integrate term by term:
\[\int\left(\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\right)dx=\ln(x+1)+\frac{2}{x+1}+C\]
Evaluate from \(1\) to \(3\):
\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]
\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]
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