int {frac{{{x^2} + 2}}{{{x^4} + 4}}} ,dx is e | vedclass

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Question

(D) To evaluate $I = \int \frac{x^2 + 2}{x^4 + 4} dx$,divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{2}{x^2}}{x^2 + \frac{4}

Vedclass · Accepted Answer

$\frac{1}{2} 	an^{-1} \left( \frac{x^2 - 2}{2x} ight) + C$

Vedclass · Answer

(D) To evaluate $I = \int \frac{x^2 + 2}{x^4 + 4} dx$,divide the numerator and denominator by $x^2$: $I = \int \frac{1 + \frac{2}{x^2}}{x^2 + \frac{4}{x^2}} dx$ $= \int \frac{1 + \frac{2}{x^2}}{(x^2 + \frac{4}{x^2})} dx$ $= \int \frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4} dx$ Let $t = x - \frac{2}{x}$. Then $dt = (1 + \frac{2}{x^2}) dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + 2^2}$ Using the formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} 	an^{-1} (\frac{x}{a}) + C$: $I = \frac{1}{2} 	an^{-1} (\frac{t}{2}) + C$ Substituting $t = x - \frac{2}{x} = \frac{x^2 - 2}{x}$ back: $I = \frac{1}{2} 	an^{-1} \left( \frac{x^2 - 2}{2x} ight) + C$ Thus,the correct option is $D$.

$\int {\frac{{{x^2} + 2}}{{{x^4} + 4}}} \,dx$ is equal to

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