Evaluate the definite integral int_{0}^{1} fr | vedclass

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(A) To evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$,we use the method of substitution.Let $x^{2}+1 = t$. Then,differentiating both sides

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

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(A) To evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$,we use the method of substitution.Let $x^{2}+1 = t$. Then,differentiating both sides with respect to $x$,we get $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.Now,change the limits of integration:When $x = 0$,$t = 0^{2} + 1 = 1$.When $x = 1$,$t = 1^{2} + 1 = 2$.Substituting these into the integral:$\int_{0}^{1} \frac{x}{x^{2}+1} d x = \int_{1}^{2} \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int_{1}^{2} \frac{1}{t} dt$.The integral of $\frac{1}{t}$ is $\log |t|$.$= \frac{1}{2} [\log |t|]_{1}^{2}$.$= \frac{1}{2} (\log 2 - \log 1)$.Since $\log 1 = 0$,we get:$= \frac{1}{2} \log 2$.

Evaluate the definite integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$.

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