The problem asks us to evaluate the integral $int frac{x^2}{x^6+1} dx$. We can try a substitution. Let $u = x^3$. Then, the derivative of $u$ with respect to $x$ is $frac{du}{dx} = 3x^2$. So, $du = 3x^2 dx$. This means $x^2 dx = frac{1}{3} du$. Now, lets substitute this into the integral: The numerator $x^2 dx$ becomes $frac{1}{3} du$. The denominator $x^6+1$ can be rewritten as $(x^3)^2+1$, which becomes $u^2+1$. So the integral transforms to: $int frac{1}{u^2+1} cdot frac{1}{3} du$ We can pull the constant $frac{1}{3}$ out of the integral: $frac{1}{3} int frac{1}{u^2+1} du$ This is a standard integral. We know that $int frac{1}{y^2+1} dy = arctan(y) + C$. Therefore, $frac{1}{3} int frac{1}{u^2+1} du = frac{1}{3} arctan(u) + C$ Finally, we substitute back $u = x^3$: $frac{1
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The problem asks us to evaluate the integral $int frac{x^2}{x^6+1} dx$. We can try a substitution. Let $u = x^3$. Then, the derivative of $u$ with respect to $x$ is $frac{du}{dx} = 3x^2$. So, $du = 3x^2 dx$. This means $x^2 dx = frac{1}{3} du$. Now, lets substitute this into the integral: The numerator $x^2 dx$ becomes $frac{1}{3} du$. The denominator $x^6+1$ can be rewritten as $(x^3)^2+1$, which becomes $u^2+1$. So the integral transforms to: $int frac{1}{u^2+1} cdot frac{1}{3} du$ We can pull the constant $frac{1}{3}$ out of the integral: $frac{1}{3} int frac{1}{u^2+1} du$ This is a standard integral. We know that $int frac{1}{y^2+1} dy = arctan(y) + C$. Therefore, $frac{1}{3} int frac{1}{u^2+1} du = frac{1}{3} arctan(u) + C$ Finally, we substitute back $u = x^3$: $frac{1
