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