int frac{x^5}{x^2+1} , dx = | vedclass

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Question

(B) Let $I = \int \frac{x^5}{x^2+1} \, dx$. We can rewrite the integrand as: $\frac{x^5}{x^2+1} = \frac{x^3(x^2+1) - x^3}{x^2+1} = x^3 - \frac{x^3}{x^

Vedclass · Accepted Answer

$\frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2+1) + c$

Vedclass · Answer

(B) Let $I = \int \frac{x^5}{x^2+1} \, dx$. We can rewrite the integrand as: $\frac{x^5}{x^2+1} = \frac{x^3(x^2+1) - x^3}{x^2+1} = x^3 - \frac{x^3}{x^2+1}$. Further,$\frac{x^3}{x^2+1} = \frac{x(x^2+1) - x}{x^2+1} = x - \frac{x}{x^2+1}$. So,$\frac{x^5}{x^2+1} = x^3 - x + \frac{x}{x^2+1}$. Now,integrating term by term: $I = \int (x^3 - x + \frac{x}{x^2+1}) \, dx = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \int \frac{2x}{x^2+1} \, dx$. Using the substitution $u = x^2+1$,$du = 2x \, dx$,we get: $I = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2+1) + c$.

$\int \frac{x^5}{x^2+1} \, dx =$

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