The integral $\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x$ is equal to
The integral $\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x$ is equal to
- $\log \frac{\left|x^3+1\right|}{x^2}+\mathrm{c}$, (where c is a constant of integration)
- $\frac{1}{2} \log \frac{\left(x^3+1\right)^2}{\left|x^3\right|}+\mathrm{c}$; (where c is a constant of integration)
- $\log \left|\frac{x^3+1}{x}\right|+\mathrm{c}$, (where c is a constant of integration)
- $\frac{1}{2} \log \frac{\left|x^3+1\right|}{x^2}+\mathrm{c}$, (where c is a constant of integration)
Solution
Let $I=\int \frac{2 x^3-1}{x^4+x} \mathrm{~d} x=\int \frac{\left(2 x-\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x}\right)} \mathrm{d} x$
$\ldots$.[Dividing $\mathrm{N}^{\mathrm{r}}$ and $\mathrm{D}^{\mathrm{r}}$ by $x^2$ ]
Put $x^2+\frac{1}{x}=\mathrm{t} \Rightarrow\left(2 x-\frac{1}{x^2}\right) \mathrm{d} x=\mathrm{dt}$
$\begin{aligned}
\therefore \quad I & =\int \frac{d t}{t} \\
& =\log |t|+c \\
& =\log \left|x^2+\frac{1}{x}\right|+c=\log \left|\frac{x^3+1}{x}\right|+c
\end{aligned}$
Asked in: MHT CET 2024 (16 May Shift 1)
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