Integrate the function: frac{3x^{2}}{x^{6}+1} | vedclass

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(A) Let $x^{3} = t$. Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$. Substituting these into the integral,we have: $\int

Vedclass · Accepted Answer

$	an^{-1}(x^{3}) + C$

Vedclass · Answer

(A) Let $x^{3} = t$. Then,differentiating both sides with respect to $x$,we get $3x^{2} dx = dt$. Substituting these into the integral,we have: $\int \frac{3x^{2}}{x^{6}+1} dx = \int \frac{dt}{t^{2}+1}$. Using the standard integral formula $\int \frac{1}{x^{2}+1} dx = 	an^{-1}(x) + C$,we get: $= 	an^{-1}(t) + C$. Substituting back $t = x^{3}$,the final result is: $= 	an^{-1}(x^{3}) + C$,where $C$ is an arbitrary constant.

Integrate the function: $\frac{3x^{2}}{x^{6}+1}$

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