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