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

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(D) Let $I = \int \frac{x^3-1}{x^3+x} dx$. Dividing the numerator by the denominator,we get: $I = \int \left(1 - \frac{x+1}{x^3+x}\right) dx = \int 1

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$x-\log |x|+\frac{1}{2} \log (x^2+1)-	an ^{-1}(x)+c$

Vedclass · Answer

(D) Let $I = \int \frac{x^3-1}{x^3+x} dx$. Dividing the numerator by the denominator,we get: $I = \int \left(1 - \frac{x+1}{x^3+x}ight) dx = \int 1 dx - \int \frac{x+1}{x(x^2+1)} dx$. Using partial fractions for $\frac{x+1}{x(x^2+1)}$: $\frac{x+1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$. $x+1 = A(x^2+1) + (Bx+C)x = (A+B)x^2 + Cx + A$. Comparing coefficients,we get $A=1$,$C=1$,and $A+B=0 \Rightarrow B=-1$. So,$\frac{x+1}{x(x^2+1)} = \frac{1}{x} + \frac{-x+1}{x^2+1} = \frac{1}{x} - \frac{x}{x^2+1} + \frac{1}{x^2+1}$. Substituting this back into the integral: $I = x - \int \left(\frac{1}{x} - \frac{x}{x^2+1} + \frac{1}{x^2+1}ight) dx$. $I = x - \log |x| + \frac{1}{2} \log (x^2+1) - 	an^{-1}(x) + c$.

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

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