KCET · Maths · Indefinite Integration
\(\int \frac{\mathrm{dx}}{\mathrm{x}^2\left(\mathrm{x}^4+1\right)^{3 / 4}}\) equals
- A \(\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c\)
- B \(\left(x^4+1\right)^{\frac{1}{4}}+c\)
- C \(-\left(\mathrm{x}^4+1\right)^{\frac{1}{4}}+\mathrm{c}\)
- D \(-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c\)
Answer & Solution
Correct Answer
(D) \(-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \int \frac{\mathrm{dx}}{\mathrm{x}^2\left(\mathrm{x}^4+1\right)^{3 / 4}} \\
& =\int \frac{\mathrm{dx}}{\mathrm{x}^5\left(1+\frac{1}{\mathrm{X}^4}\right)^{3 / 4}} \quad 1+\frac{1}{\mathrm{x}^4}=\mathrm{T} \\
& -4 \mathrm{x}^{-5} \mathrm{dx}=\mathrm{dt} \\
& =-\frac{1}{4} \int \mathrm{t}^{-3 / 4} \mathrm{dt}=-\frac{1}{4} \frac{\mathrm{t}^{1 / 4}}{\frac{1}{4}}+\mathrm{c}=-\left(1+\frac{1}{\mathrm{x}^4}\right)^{1 / 4}
\end{aligned}\)
& \int \frac{\mathrm{dx}}{\mathrm{x}^2\left(\mathrm{x}^4+1\right)^{3 / 4}} \\
& =\int \frac{\mathrm{dx}}{\mathrm{x}^5\left(1+\frac{1}{\mathrm{X}^4}\right)^{3 / 4}} \quad 1+\frac{1}{\mathrm{x}^4}=\mathrm{T} \\
& -4 \mathrm{x}^{-5} \mathrm{dx}=\mathrm{dt} \\
& =-\frac{1}{4} \int \mathrm{t}^{-3 / 4} \mathrm{dt}=-\frac{1}{4} \frac{\mathrm{t}^{1 / 4}}{\frac{1}{4}}+\mathrm{c}=-\left(1+\frac{1}{\mathrm{x}^4}\right)^{1 / 4}
\end{aligned}\)
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