int_{2}^{3} frac{x}{x^{2}-1} d x= | vedclass

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(B) To evaluate the integral $I = \int_{2}^{3} \frac{x}{x^{2}-1} d x$,let $u = x^{2}-1$. Then $du = 2x \, dx$,which implies $x \, dx = \frac{1}{2} du$

Vedclass · Accepted Answer

$\left(\frac{1}{2}\right) \log \left(\frac{8}{3}\right)$

Vedclass · Answer

(B) To evaluate the integral $I = \int_{2}^{3} \frac{x}{x^{2}-1} d x$,let $u = x^{2}-1$. Then $du = 2x \, dx$,which implies $x \, dx = \frac{1}{2} du$. When $x = 2$,$u = 2^{2}-1 = 3$. When $x = 3$,$u = 3^{2}-1 = 8$. Substituting these into the integral: $I = \int_{3}^{8} \frac{1}{u} \cdot \frac{1}{2} du$ $I = \frac{1}{2} [\log |u|]_{3}^{8}$ $I = \frac{1}{2} (\log 8 - \log 3)$ $I = \frac{1}{2} \log \left(\frac{8}{3}\right)$.

$\int_{2}^{3} \frac{x}{x^{2}-1} d x=$

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