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\( \int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} \mathrm{dx} \) is equal to
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The correct answer is:
\( -\frac{\left(1+x^{4}\right)^{1 / 4}}{x}+C \)
Given that \( I=\int \frac{d x}{x^{2}\left(x^{4}+1\right)^{3 / 4}} \)
\( =\int \frac{d x}{x^{2}\left[x^{4}\left(1+\frac{1}{x^{4}}\right)\right]^{3 / 4}} \)
\( =\int \frac{d x}{x^{2} x^{3}\left(1+x^{-4}\right)^{3 / 4}}=\int \frac{x^{-5}}{\left(1+x^{-4}\right)^{3 / 4}} d x \)
Let \( 1+x^{-4}=t^{4} \) then \( -4 x^{5} d x=4 t^{3} d t \)
So, \( I=-\frac{1}{4} \int \frac{4 t^{3}}{t^{3}} d t=-\int d t=-t \)
\( =-\left(1+\frac{1}{x^{4}}\right)^{1 / 4}=-\left(\frac{1+x^{4}}{x^{4}}\right)^{1 / 4}=-\frac{\left(1+x^{4}\right)^{1 / 4}}{x}+C \)
\( =\int \frac{d x}{x^{2}\left[x^{4}\left(1+\frac{1}{x^{4}}\right)\right]^{3 / 4}} \)
\( =\int \frac{d x}{x^{2} x^{3}\left(1+x^{-4}\right)^{3 / 4}}=\int \frac{x^{-5}}{\left(1+x^{-4}\right)^{3 / 4}} d x \)
Let \( 1+x^{-4}=t^{4} \) then \( -4 x^{5} d x=4 t^{3} d t \)
So, \( I=-\frac{1}{4} \int \frac{4 t^{3}}{t^{3}} d t=-\int d t=-t \)
\( =-\left(1+\frac{1}{x^{4}}\right)^{1 / 4}=-\left(\frac{1+x^{4}}{x^{4}}\right)^{1 / 4}=-\frac{\left(1+x^{4}\right)^{1 / 4}}{x}+C \)
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