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The integral $\int_{}^{} \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$ equals

$\left[ \frac{x^{4}+1}{x^{4}}\right]^{\frac{1}{4}}+C$

$(x^{4}+1)^{\frac{1}{4}}+C$

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

$-\left[ \frac{x^{4}+1}{x^{4}}\right]^{\frac{1}{4}}+C$

Option D

$\int_{}^{} \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$ = $\int_{}^{} \frac{dx}{x^{5}(1+\frac{1}{x^{4}})^{\frac{3}{4}}}$

Put $1+\frac{1}{x^{4}}=t^{4}$

$\Rightarrow$ $-\frac{4}{x^{5}}dx=4t^{3}dt$

$\Rightarrow$ $\frac{dx}{x^{5}}=-t^{3}dt=\int_{}^{} \frac{-t^{3}dt}{t^{3}}$

= $-\int_{}^{} dt=-t+C=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$

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