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Answer: \(\frac{1}{4}\log\frac{9}{7}\)

\(x^2+6x+5=(x+3)^2-2^2\). Using \(\int\frac{dx}{X^2-a^2}=\frac{1}{2a}\log\left|\frac{X-a}{X+a}\right|\):

\(I=\frac{1}{4}\left[\log\left|\frac{x+3-2}{x+3+2}\right|\right]_1^2=\frac{1}{4}\left[\log\frac{x+1}{x+5}\right]_1^2\)

\(=\frac{1}{4}\left[\log\frac{3}{7}-\log\frac{2}{6}\right]=\frac{1}{4}\log\left(\frac{3}{7}\times3\right)=\frac{1}{4}\log\frac{9}{7}\)

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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