int frac{2x tan^{-1}(x^2)}{1 + x^4} , dx = | vedclass

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Question

(B) Let $t = 	an^{-1}(x^2)$.Then,differentiating both sides with respect to $x$,we get $dt = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \, dx = \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $t = 	an^{-1}(x^2)$.Then,differentiating both sides with respect to $x$,we get $dt = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) \, dx = \frac{2x}{1 + x^4} \, dx$.Substituting these into the integral,we have:$\int 	an^{-1}(x^2) \cdot \frac{2x}{1 + x^4} \, dx = \int t \, dt$.Integrating with respect to $t$,we get $\frac{t^2}{2} + c$.Substituting back the value of $t$,we obtain $\frac{1}{2} (	an^{-1}(x^2))^2 + c$.

$\int \frac{2x \tan^{-1}(x^2)}{1 + x^4} \, dx = $

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