int frac{x}{1 + x^4} , dx = | vedclass

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(B) Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$. Substitutin

Vedclass · Accepted Answer

$\frac{1}{2} 	an^{-1}(x^2) + c$

Vedclass · Answer

(B) Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$. Substituting these into the integral: $\int \frac{x}{1 + x^4} \, dx = \int \frac{1}{1 + (x^2)^2} \cdot (x \, dx) = \int \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt$. $= \frac{1}{2} \int \frac{1}{1 + t^2} \, dt$. Using the standard integral formula $\int \frac{1}{1 + t^2} \, dt = 	an^{-1}(t) + c$,we get: $= \frac{1}{2} 	an^{-1}(t) + c$. Substituting $t = x^2$ back,the final result is $\frac{1}{2} 	an^{-1}(x^2) + c$.

$\int \frac{x}{1 + x^4} \, dx = $

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