Integral of d x x² [1+x⁴ ]^ 3 / 4 .

Step-by-step solution

Let I = dx x^2(1+x^4)^ 3/4 . Factor out x^4 from the term in the bracket: I = dx x^2 [x^4(x⁻⁴ + 1)]^ 3/4 = dx x^2 x^3 (x⁻⁴ + 1)^ 3/4 = dx x^5 (x⁻⁴ + 1)^ 3/4 . Let u = x⁻⁴ + 1 . Then du = -4x⁻⁵ dx , which implies x⁻⁵ dx = - 1 4 du . Substituting these into the integral: I = - 1 4 u^ -3/4 du . Integrating with respect to u : I = - 1 4 ( u^ 1/4 1/4 ) + C = -u^ 1/4 + C . Substituting back u = x⁻⁴ + 1 = 1+x^4 x^4 : I = - ( 1+x^4 x^4 )^ 1/4 + C = - (1+x^4)^ 1/4 x + C . Comparing this result with the provided options, non