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

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(B) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$.We can rewrite the denominator as $x(x^3+1)$.So,$I = \int \frac{2 x^3-1}{x(x^3+1)} \,d x$.Using partial

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$\log \frac{\left(x^3+1\right)}{x}+C$

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(B) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$.We can rewrite the denominator as $x(x^3+1)$.So,$I = \int \frac{2 x^3-1}{x(x^3+1)} \,d x$.Using partial fractions or observation,we can write $\frac{2x^3-1}{x(x^3+1)} = \frac{A}{x} + \frac{B x^2}{x^3+1}$.By inspection,$\frac{2x^3-1}{x(x^3+1)} = \frac{-1}{x} + \frac{3x^2}{x^3+1}$.Integrating both terms:$I = \int \left( \frac{3x^2}{x^3+1} - \frac{1}{x} \right) \,d x$.$I = \log|x^3+1| - \log|x| + C$.$I = \log \left| \frac{x^3+1}{x} \right| + C$.

The value of $\int \frac{2 x^3-1}{x^4+x} \,d x$ is equal to (where $C$ is a constant of integration.)

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