int frac{x^{n-1}}{x^{2n} + 4} dx = | vedclass

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(A) Let $I = \int \frac{x^{n-1}}{x^{2n} + 4} dx$. Substitute $x^n = t$. Then,differentiating both sides with respect to $x$,we get $n x^{n-1} dx = dt$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x^{n-1}}{x^{2n} + 4} dx$. Substitute $x^n = t$. Then,differentiating both sides with respect to $x$,we get $n x^{n-1} dx = dt$,which implies $x^{n-1} dx = \frac{1}{n} dt$. Substituting these into the integral,we get: $I = \frac{1}{n} \int \frac{dt}{t^2 + 2^2}$. Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) + c$,we have: $I = \frac{1}{n} \cdot \frac{1}{2} 	an^{-1} \left( \frac{t}{2} ight) + c$. Substituting $t = x^n$ back,we get: $I = \frac{1}{2n} 	an^{-1} \left( \frac{x^n}{2} ight) + c$.

$\int \frac{x^{n-1}}{x^{2n} + 4} dx =$

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