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Integral of $\int \frac{d x}{x^{2}\left[1+x^{4}\right]^{3 / 4}}$.
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Let $I=\int \frac{d x}{x^{2}\left[1+x^{4}\right]^{\frac{3}{4}}}=\int \frac{d x}{x^{2}\left(x^{4}\left(\frac{1}{x^{4}}+1\right)\right)^{3 / 4}}$ [taking $x^{4}$ common from denominator]
$$
\Rightarrow \int \frac{d x}{x^{5}\left(\frac{1}{x^{4}}+1\right)^{3 / 4}}
$$
Let $\left(\frac{1}{x^{4}}+1\right)=t^{4} \Rightarrow \frac{-4 d x}{x^{5}}=4 t^{3} d t \Rightarrow \frac{d x}{x^{5}}=-t^{3} d t$ $\therefore I=\int \frac{-t^{3} d t}{t^{3}}=-t+C=-\left(\frac{1}{x^{4}}+1\right)^{1 / 4}+C$
$$
\Rightarrow \int \frac{d x}{x^{5}\left(\frac{1}{x^{4}}+1\right)^{3 / 4}}
$$
Let $\left(\frac{1}{x^{4}}+1\right)=t^{4} \Rightarrow \frac{-4 d x}{x^{5}}=4 t^{3} d t \Rightarrow \frac{d x}{x^{5}}=-t^{3} d t$ $\therefore I=\int \frac{-t^{3} d t}{t^{3}}=-t+C=-\left(\frac{1}{x^{4}}+1\right)^{1 / 4}+C$
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