A primitive of f(x) = frac{x}{1 + x^2} is: | vedclass

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(C) To find the primitive (indefinite integral) of $f(x) = \frac{x}{1 + x^2}$,we evaluate the integral $I = \int \frac{x}{1 + x^2} dx$.Let $t = 1 + x^

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$\frac{\log_e(x^2 + 1)}{2}$

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(C) To find the primitive (indefinite integral) of $f(x) = \frac{x}{1 + x^2}$,we evaluate the integral $I = \int \frac{x}{1 + x^2} dx$.Let $t = 1 + x^2$.Then,differentiating both sides with respect to $x$,we get $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{1}{t} dt$.Integrating,we obtain $I = \frac{1}{2} \log_e |t| + C$.Substituting back $t = 1 + x^2$,we get $I = \frac{1}{2} \log_e(1 + x^2) + C$.

$A$ primitive of $f(x) = \frac{x}{1 + x^2}$ is:

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