int frac{x^3+2 x}{x^4+4} d x= | vedclass

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(A) Let $I = \int \frac{x^3+2x}{x^4+4} dx$. Split the integral into two parts: $I = \int \frac{x^3}{x^4+4} dx + \int \frac{2x}{x^4+4} dx$. For the fir

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$\frac{1}{2}\left[	an ^{-1}\left(\frac{x^2}{2}ight)+\log \left(\frac{\sqrt{x^4+4}}{2}ight)ight]+C$

Vedclass · Answer

(A) Let $I = \int \frac{x^3+2x}{x^4+4} dx$. Split the integral into two parts: $I = \int \frac{x^3}{x^4+4} dx + \int \frac{2x}{x^4+4} dx$. For the first part, let $u = x^4+4$, then $du = 4x^3 dx$, so $\int \frac{x^3}{x^4+4} dx = \frac{1}{4} \log|x^4+4| + C_1$. For the second part, let $t = x^2$, then $dt = 2x dx$, so $\int \frac{2x}{(x^2)^2+4} dx = \int \frac{dt}{t^2+2^2} = \frac{1}{2} 	an^{-1}(\frac{t}{2}) + C_2 = \frac{1}{2} 	an^{-1}(\frac{x^2}{2}) + C_2$. Combining these, $I = \frac{1}{4} \log(x^4+4) + \frac{1}{2} 	an^{-1}(\frac{x^2}{2}) + C$. Note that $\frac{1}{4} \log(x^4+4) = \frac{1}{2} \log(\sqrt{x^4+4}) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2} 	imes 2) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2}) + \frac{1}{2} \log(2)$. Absorbing the constant $\frac{1}{2} \log(2)$ into $C$, we get $I = \frac{1}{2} \left[ 	an^{-1}(\frac{x^2}{2}) + \log(\frac{\sqrt{x^4+4}}{2}) ight] + C$.

$\int \frac{x^3+2 x}{x^4+4} d x=$

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