Evaluate the definite integral: frac{1}{2} in | vedclass

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Question

(A) Let $I = \frac{1}{2} \int_2^3 \frac{2x}{x^2+1} dx$.Using the substitution method, let $u = x^2+1$.Then $du = 2x dx$.When $x=2$, $u = 2^2+1 = 5$.Wh

Vedclass · Accepted Answer

$\frac{1}{2} \log(2)$

Vedclass · Answer

(A) Let $I = \frac{1}{2} \int_2^3 \frac{2x}{x^2+1} dx$.Using the substitution method, let $u = x^2+1$.Then $du = 2x dx$.When $x=2$, $u = 2^2+1 = 5$.When $x=3$, $u = 3^2+1 = 10$.Substituting these into the integral:$I = \frac{1}{2} \int_5^{10} \frac{1}{u} du$$I = \frac{1}{2} [\log |u|]_5^{10}$$I = \frac{1}{2} (\log 10 - \log 5)$$I = \frac{1}{2} \log \left(\frac{10}{5}\right)$$I = \frac{1}{2} \log(2)$.

Evaluate the definite integral: $\frac{1}{2} \int_2^3 \frac{2x}{x^2+1} dx = $

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