int frac{1}{x^{2}left(x^{4}+1right)^{3 / 4}} | vedclass

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Question

(A) Given the integral $ I = \int \frac{dx}{x^{2}(x^{4}+1)^{3/4}} $. We can factor $ x^{4} $ out of the parenthesis: $ I = \int \frac{dx}{x^{2}[x^{4}(

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the integral $ I = \int \frac{dx}{x^{2}(x^{4}+1)^{3/4}} $. We can factor $ x^{4} $ out of the parenthesis: $ I = \int \frac{dx}{x^{2}[x^{4}(1 + \frac{1}{x^{4}})]^{3/4}} $. Simplifying the expression: $ I = \int \frac{dx}{x^{2} \cdot x^{3}(1 + x^{-4})^{3/4}} = \int \frac{x^{-5}}{(1 + x^{-4})^{3/4}} dx $. Let $ 1 + x^{-4} = t^{4} $. Then,differentiating both sides with respect to $ x $,we get $ -4x^{-5} dx = 4t^{3} dt $,which implies $ x^{-5} dx = -t^{3} dt $. Substituting these into the integral: $ I = \int \frac{-t^{3} dt}{t^{3}} = -\int dt = -t + C $. Substituting back $ t = (1 + x^{-4})^{1/4} $: $ I = -(1 + x^{-4})^{1/4} + C = -(\frac{x^{4}+1}{x^{4}})^{1/4} + C = -\frac{(1+x^{4})^{1/4}}{x} + C $.

$ \int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} dx $ is equal to

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