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

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Question

(B) We are given 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 =

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We are given 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,we get: $I = \int \frac{5(x^2 + 1)(x^4 - x^2 + 1)}{x^2 + 1} dx$ $I = \int 5(x^4 - x^2 + 1) dx$ Integrating term by term: $I = 5 \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) + c$ $I = x^5 - \frac{5}{3}x^3 + 5x + c$. Thus,the correct option is $B$.

$\int \frac{5(x^6 + 1)}{x^2 + 1} dx = $

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