int frac{x^{3}+x}{x-1} d x

Question

Accepted Answer

Step:1 ：Perform polynomial long division to simplify the integrand: (frac{x^3 + x}{x - 1} = x^2 + x + 2 + frac{2}{x - 1})Step:2 ：Integrate each part separately: (int (x^2 + x + 2) dx + int frac{2}{x - 1} dx)Step:3 ：Integrate the first part: (frac{x^3}{3} + frac{x^2}{2} + 2x)Step:4 ：Integrate the second part: (2ln|x - 1|)Step:5 ：Combine the results: (boxed{frac{x^3}{3} + frac{x^2}{2} + 2x + 2ln|x - 1| + C})

Answer

Step: 1 ：Perform polynomial long division to simplify the integrand: (frac{x^3 + x}{x - 1} = x^2 + x + 2 + frac{2}{x - 1})Step: 2 ：Integrate each part separately: (int (x^2 + x + 2) dx + int frac{2}{x - 1} dx)Step: 3 ：Integrate the first part: (frac{x^3}{3} + frac{x^2}{2} + 2x)Step: 4 ：Integrate the second part: (2ln|x - 1|)Step: 5 ：Combine the results: (boxed{frac{x^3}{3} + frac{x^2}{2} + 2x + 2ln|x - 1| + C})

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

Answer

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