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

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Question

(B) We have the integral $I = \int {\frac{{{x^3} - 1}}{{{x^3} + x}}dx}$.First,simplify the integrand: $\frac{{{x^3} - 1}}{{{x^3} + x}} = \frac{{{x^3}

Vedclass · Accepted Answer

$x - \log |x| + \log \sqrt {{x^2} + 1} - {	an ^{ - 1}}x + c$

Vedclass · Answer

(B) We have the integral $I = \int {\frac{{{x^3} - 1}}{{{x^3} + x}}dx}$.First,simplify the integrand: $\frac{{{x^3} - 1}}{{{x^3} + x}} = \frac{{{x^3} + x - x - 1}}{{x({x^2} + 1)}} = \frac{{{x^3} + x}}{{x({x^2} + 1)}} - \frac{{x + 1}}{{x({x^2} + 1)}} = 1 - \frac{{x + 1}}{{x({x^2} + 1)}}$.Using partial fractions for $\frac{{x + 1}}{{x({x^2} + 1)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}}$.$x + 1 = A({x^2} + 1) + (Bx + C)x = (A + B){x^2} + Cx + A$.Comparing coefficients,we get $A = 1$,$C = 1$,and $A + B = 0 \implies B = -1$.So,$\frac{{x + 1}}{{x({x^2} + 1)}} = \frac{1}{x} - \frac{{x - 1}}{{{x^2} + 1}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}$.Now,$I = \int {\left( {1 - \left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}}ight)} ight)dx} = \int {\left( {1 - \frac{1}{x} + \frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}}} ight)dx}$.Integrating term by term: $I = x - \log |x| + \frac{1}{2}\log ({x^2} + 1) - {	an ^{ - 1}}x + c$.Since $\frac{1}{2}\log ({x^2} + 1) = \log \sqrt {{x^2} + 1}$,the result is $x - \log |x| + \log \sqrt {{x^2} + 1} - {	an ^{ - 1}}x + c$.

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

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