Integral Calculus
86482
\(\int_{2}^{3} \frac{x}{x^{2}-1} d x=\)
1 \(\left(\frac{1}{3}\right) \log \left(\frac{8}{3}\right)\)
2 \(\left(\frac{-1}{3}\right) \log \left(\frac{8}{3}\right)\)
3 \(\left(\frac{1}{2}\right) \log \left(\frac{8}{3}\right)\)
4 \(\left(\frac{-1}{2}\right) \log \left(\frac{8}{3}\right)\)
Explanation:
(C) : \(I=\int_{2}^{3} \frac{x}{x^{2}-1} d x\)
Let, \(\quad x^{2}-1 \stackrel{2}{=} \mathrm{t}\)
\(2 \mathrm{xdx}=\mathrm{dt}\)
\(\mathrm{x} . \mathrm{dx}=\frac{1}{2} \mathrm{dt}\)
When, \(x=2, t=3\)
When, \(\mathrm{x}=3, \mathrm{t}=8\)
\(\therefore \mathrm{I}=\frac{1}{2} \int_{3}^{8} \frac{1}{\mathrm{t}} \mathrm{dt}=\frac{1}{2}[\log \mathrm{t}]_{3}^{8}=\frac{1}{2}[\log 8-\log 3]=\frac{1}{2} \log \left[\frac{8}{3}\right]\)