The integral int frac{2 x^3-1}{x^4+x} ,d x is | vedclass

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Question

(C) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{2 x - \frac{1}{x^2}}{x^2 + \frac{1}{x}}

Vedclass · Accepted Answer

$\log \left|\frac{x^3+1}{x}\right|+c$,(where $c$ is a constant of integration)

Vedclass · Answer

(C) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{2 x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \,d x$. Let $t = x^2 + \frac{1}{x}$. Then $dt = (2x - \frac{1}{x^2}) \,d x$. Substituting these into the integral: $I = \int \frac{dt}{t} = \log |t| + c$. Substituting back $t = x^2 + \frac{1}{x} = \frac{x^3+1}{x}$: $I = \log \left| \frac{x^3+1}{x} \right| + c$.

The integral $\int \frac{2 x^3-1}{x^4+x} \,d x$ is equal to

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