int (x + frac{1}{x})^3 dx = | vedclass

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(B) We expand the integrand using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ or $a^3 + 3a^2b + 3ab^2 + b^3$.$\int (x + \frac{1}{x})^3 dx = \int (x^

Vedclass · Accepted Answer

$\frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| - \frac{1}{2x^2} + c$

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(B) We expand the integrand using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ or $a^3 + 3a^2b + 3ab^2 + b^3$.$\int (x + \frac{1}{x})^3 dx = \int (x^3 + 3x^2(\frac{1}{x}) + 3x(\frac{1}{x^2}) + \frac{1}{x^3}) dx$$= \int (x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}) dx$Integrating term by term:$= \frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| + \frac{x^{-2}}{-2} + c$$= \frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| - \frac{1}{2x^2} + c$

$\int (x + \frac{1}{x})^3 dx = $

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