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

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(B) We have the integral $I = \int \frac{x+1}{x^3-1} \, dx$. Using the factorization $x^3-1 = (x-1)(x^2+x+1)$,we write the integrand as partial fracti

Vedclass · Accepted Answer

$\frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right)+c$

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(B) We have the integral $I = \int \frac{x+1}{x^3-1} \, dx$. Using the factorization $x^3-1 = (x-1)(x^2+x+1)$,we write the integrand as partial fractions: $\frac{x+1}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}$. Equating numerators: $x+1 = A(x^2+x+1) + (Bx+C)(x-1)$. For $x=1$: $2 = A(3) \implies A = \frac{2}{3}$. Comparing coefficients of $x^2$: $0 = A+B \implies B = -\frac{2}{3}$. Comparing constants: $1 = A-C \implies C = A-1 = \frac{2}{3}-1 = -\frac{1}{3}$. Thus,$I = \int \left( \frac{2/3}{x-1} + \frac{-2/3x - 1/3}{x^2+x+1} \right) \, dx = \frac{2}{3} \ln|x-1| - \frac{1}{3} \int \frac{2x+1}{x^2+x+1} \, dx$. $I = \frac{2}{3} \ln|x-1| - \frac{1}{3} \ln|x^2+x+1| + c = \frac{1}{3} \ln \left( \frac{(x-1)^2}{x^2+x+1} \right) + c$.

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

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