If $\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$, then I is
If $\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$, then I is
- $\left(\frac{x^4+1}{x}\right)^{\frac{1}{4}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\frac{\left(x^4-1\right)^{\frac{1}{4}}}{x}+c$, where $\mathrm{c}$ is a constant of integration.
- $-\frac{\left(x^4+1\right)^{\frac{1}{4}}}{x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $-\left(\frac{x^4+1}{x}\right)^{\frac{1}{4}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
Let $\mathrm{I}=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \mathrm{~d} x=\int \frac{\mathrm{d} x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}}$
Put $1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{dx}=\mathrm{dt}$
$\begin{aligned}
\therefore \quad I & =-\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^{\frac{3}{4}}}=-\frac{1}{4} \times 4 \mathrm{t}^{\frac{1}{4}}+\mathrm{c}=-\mathrm{t}^{\frac{1}{4}}+\mathrm{c} \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c} \\
& =\frac{-\left(x^4+1\right)^{\frac{1}{4}}}{x}+\mathrm{c}
\end{aligned}$
Asked in: MHT CET 2023 (10 May Shift 1)
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