\(\int\displaystyle\frac{dx}{x^3\sqrt{(1+x^4)}} =\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2x^2}\sqrt{1+x^4}+C\), C is an arbitrary constant
Taking \(x^2\) out of the root (\(x^2 \gt 0\)):
\(\displaystyle \sqrt{1 + x^4} = x^2\sqrt{1 + \frac{1}{x^4}}\)
\(\displaystyle \int \frac{dx}{x^3\sqrt{1 + x^4}} = \int \frac{dx}{x^5\sqrt{1 + x^{-4}}}\)
Now let (substitution)
\(t = 1 + x^{-4}\)
\(\displaystyle dt = -4x^{-5}\,dx\), so \(\displaystyle \frac{dx}{x^5} = -\frac{1}{4}\,dt\)
Then
\(\displaystyle \int \frac{dx}{x^5\sqrt{1 + x^{-4}}} = -\frac{1}{4}\int t^{-1/2}\,dt\)
\(\displaystyle = -\frac{1}{2}\sqrt{t} + C\)
\(\displaystyle = -\frac{1}{2}\sqrt{\frac{x^4 + 1}{x^4}} + C\)
\(\displaystyle = -\frac{1}{2x^2}\sqrt{1 + x^4} + C\)