int frac{5(x^6+1)}{x^2+1} , dx = (Where C is | vedclass

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(B) We have the integral $I = \int \frac{5(x^6+1)}{x^2+1} \, dx$. Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,where $a=x^2$ and $b=1$,w

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$x^5-\frac{5x^3}{3}+5x+C$

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(B) We have the integral $I = \int \frac{5(x^6+1)}{x^2+1} \, dx$. Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,where $a=x^2$ and $b=1$,we can write $x^6+1 = (x^2)^3+1^3 = (x^2+1)(x^4-x^2+1)$. Substituting this into the integral: $I = \int \frac{5(x^2+1)(x^4-x^2+1)}{x^2+1} \, dx$ $I = 5 \int (x^4-x^2+1) \, dx$ Integrating term by term: $I = 5 \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) + C$ $I = x^5 - \frac{5x^3}{3} + 5x + C$.

$\int \frac{5(x^6+1)}{x^2+1} \, dx =$ (Where $C$ is a constant of integration.)

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