MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\) is
- A \(\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- B \(\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- C \(-\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- D \(-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Answer & Solution
Correct Answer
(D) \(-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
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 \(1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad I & =-\frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}} \\
& =-\frac{1}{4} \times 4 t^{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+c=-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c
\end{aligned}\)
Put \(1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad I & =-\frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}} \\
& =-\frac{1}{4} \times 4 t^{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+c=-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c
\end{aligned}\)
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