Evaluate the integral: int frac{1}{left(x+fra | vedclass

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(A) Let $I = \int \frac{d x}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x d x}{\lef

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$\frac{1}{2} \sec ^{-1}\left(x^2+2\right)+c$

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(A) Let $I = \int \frac{d x}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x d x}{\left(x^2+2\right) \sqrt{x^4+4 x^2+3}}$.Rewrite the term inside the square root: $x^4+4 x^2+3 = (x^2+2)^2 - 1$.So, $I = \int \frac{x d x}{\left(x^2+2\right) \sqrt{(x^2+2)^2 - 1}}$.Let $t = x^2+2$, then $dt = 2x dx$, which implies $x dx = \frac{1}{2} dt$.Substituting these into the integral:$I = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2-1}}$.Using the standard integral formula $\int \frac{dt}{t \sqrt{t^2-1}} = \sec^{-1}(t) + C$:$I = \frac{1}{2} \sec^{-1}(t) + C$.Substituting back $t = x^2+2$:$I = \frac{1}{2} \sec^{-1}(x^2+2) + C$.

Evaluate the integral: $\int \frac{1}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} d x$

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