∫frac1x^2≤ft(x^4+1)^3/4 dx is equal to:

The correct answer is Option (2) → \(-\left(1+\displaystyle\frac{1}{x^4}\right)^{\frac{1}{4}}+C\)

Taking \(x^4\) out of the bracket (for x > 0):

    \( \displaystyle \left(x^4+1\right)^{3/4} = x^3\left(1+\frac{1}{x^4}\right)^{3/4} \)

So

    \( \displaystyle \int\frac{dx}{x^2\left(x^4+1\right)^{3/4}} = \int\frac{dx}{x^5\left(1+x^{-4}\right)^{3/4}} \)

Now use the substitution \(t = 1 + x^{-4}\):

    \( \displaystyle dt = -4x^{-5}\,dx, \quad \frac{dx}{x^5} = -\frac{dt}{4} \)

Then

    \( \displaystyle \int\frac{dx}{x^5\left(1+x^{-4}\right)^{3/4}} = -\frac{1}{4}\int t^{-3/4}\,dt \)

    \( \displaystyle = -\frac{1}{4}\times\frac{t^{1/4}}{1/4} + C = -t^{1/4} + C \)

    \( \displaystyle = -\left(1+\frac{1}{x^4}\right)^{1/4} + C \)

NCERT (2026-27 edition):
Integrals