The value of $\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x$ is equal to (where $C$ is a constant of integration.)
The value of $\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x$ is equal to (where $C$ is a constant of integration.)
- $\frac{1}{2} \log \frac{\left(x^3+1\right)^2}{x^3}+C$
- $\log \frac{\left(x^3+1\right)}{x}+C$
- $\log \left(\frac{x^3+1}{x^2}\right)+C$
- $\frac{1}{2} \log \frac{\left(x^3+1\right)}{x^2}+C$
Solution
$\begin{aligned} & \int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x=\int \frac{2 x^3-1}{x\left(x^3+1\right)} \mathrm{d} x=\int\left(\frac{3 x^2}{x^3+1}-\frac{1}{x}\right) \mathrm{d} x \\ & =\log \left(x^3+1\right)-\log x+c \\ & =\log \left(\frac{x^3+1}{x}\right)+c\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 2)
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