int frac{x e^{left(frac{x^2}{x^2-2}right)}}{x | vedclass

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(A) Let $I = \int \frac{x e^{\left(\frac{x^2}{x^2-2}\right)}}{x^4-4 x^2+4} d x$. Note that $x^4-4x^2+4 = (x^2-2)^2$. So, $I = \int \frac{x e^{\left(\f

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$\frac{-1}{4} e^{\frac{x^2}{x^2-2}}+C$

Vedclass · Answer

(A) Let $I = \int \frac{x e^{\left(\frac{x^2}{x^2-2}\right)}}{x^4-4 x^2+4} d x$. Note that $x^4-4x^2+4 = (x^2-2)^2$. So, $I = \int \frac{x e^{\left(\frac{x^2}{x^2-2}\right)}}{(x^2-2)^2} d x$. Let $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{1}{2} dt$. Substituting these into the integral: $I = \frac{1}{2} \int \frac{e^{\left(\frac{t}{t-2}\right)}}{(t-2)^2} dt$. Now, let $p = \frac{t}{t-2}$. Then $dp = \frac{(t-2)(1) - t(1)}{(t-2)^2} dt = \frac{-2}{(t-2)^2} dt$. This implies $\frac{1}{(t-2)^2} dt = -\frac{1}{2} dp$. Substituting into the integral: $I = \frac{1}{2} \int e^p \left(-\frac{1}{2} dp\right) = -\frac{1}{4} \int e^p dp = -\frac{1}{4} e^p + C$. Substituting back $p = \frac{x^2}{x^2-2}$: $I = -\frac{1}{4} e^{\frac{x^2}{x^2-2}} + C$.

$\int \frac{x e^{\left(\frac{x^2}{x^2-2}\right)}}{x^4-4 x^2+4} d x=$

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