Evaluate the integral: int frac{x^3 , dx}{1+x | vedclass

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(B) To evaluate the integral $I = \int \frac{x^3 \, dx}{1+x^8}$,we can rewrite the denominator as $1 + (x^4)^2$.Let $u = x^4$.Then,differentiating bot

Vedclass · Accepted Answer

$\frac{1}{4} 	an^{-1} x^4 + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{x^3 \, dx}{1+x^8}$,we can rewrite the denominator as $1 + (x^4)^2$.Let $u = x^4$.Then,differentiating both sides with respect to $x$,we get $du = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{du}{4}$.Substituting these into the integral,we get:$I = \int \frac{du/4}{1+u^2} = \frac{1}{4} \int \frac{du}{1+u^2}$.Using the standard integral formula $\int \frac{du}{1+u^2} = 	an^{-1}(u) + c$,we obtain:$I = \frac{1}{4} 	an^{-1}(u) + c$.Substituting $u = x^4$ back,we get the final result:$I = \frac{1}{4} 	an^{-1}(x^4) + c$.

Evaluate the integral: $\int \frac{x^3 \, dx}{1+x^8}$

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