frac{1}{2} int_2^3 frac{2 x}{x^2+1} d x= . . | vedclass

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Question

(B) 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 $x=2$,$u = 2^2+1 = 5$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) 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: $I = \frac{1}{2} \int_5^{10} \frac{1}{u} du$ $I = \frac{1}{2} [\log |u|]_5^{10}$ $I = \frac{1}{2} (\log 10 - \log 5)$ $I = \frac{1}{2} \log \left(\frac{10}{5}\right)$ $I = \frac{1}{2} \log (2)$.

$\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x=$ . . . . . . .

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