The value of $\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is
The value of $\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is
- $-\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}$, (where c is a constant of integration)
- $\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}$, (where c is a constant of integration)
- $\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}$, (where c is a constant of integration)
- $-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}$, (where c is a constant of integration)
Solution
$\begin{aligned}
& \text { Let } \mathrm{I}=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \mathrm{~d} x=\int \frac{\mathrm{d} x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} \\
& \text { Put } 1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=-\frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}}=-\frac{1}{4} \times 4 t^{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+c
\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 2)
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