CUET · MATHS · PYQ PAPER 2025
\(\int \frac{d x}{x^3 \sqrt{\left(1+x^4\right)}}=\)
- A \(\frac{1}{4 x^2} \sqrt{1+x^4}+C\), where \(C\) is an arbitrary constant
- B \(\frac{1}{x^2} \sqrt{1+x^4}+C\), where \(C\) is an arbitrary constant
- C \(\frac{1}{2 x^2} \sqrt{1+x^4}+C\), where \(C\) is an arbitrary constant
- D \(-\frac{1}{2 x^2} \sqrt{1+x^4}+C\), where \(C\) is an arbitrary constant
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2 x^2} \sqrt{1+x^4}+C\), where \(C\) is an arbitrary constant
Step-by-step Solution
Detailed explanation
Let \(u = 1+x^{-4}\). Then \(du = -4x^{-5} dx \Rightarrow x^{-5} dx = -\frac{1}{4} du\). \(\int \frac{d x}{x^3 \sqrt{\left(1+x^4\right)}} = \int \frac{x^{-5} dx}{\sqrt{1+x^{-4}}} = \int \frac{-\frac{1}{4} du}{\sqrt{u}}\)…
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