Question

\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]

Answer 1

\[\int\frac{dx}{x^2 \left( x^4 + 1 \right)^\frac{3}{4}}\]
\[ = \int\frac{dx}{x^2 \left[ x^4 \left( 1 + \frac{1}{x^4} \right) \right]^\frac{3}{4}}\]
\[ = \int \frac{dx}{x^2 . x^3 \left( 1 + \frac{1}{x^4} \right)^\frac{3}{4}}\]
\[ = \int\frac{\left( 1 + \frac{1}{x^4} \right)^{- \frac{3}{4}}}{x^5}  \text{ dx }\]
\[\text{Let 1 }+ \frac{1}{x^4} = t\]
\[ \Rightarrow - \frac{4}{x^5}dx = \text{ dt }\]
\[ \Rightarrow \frac{dx}{x^5} = - \frac{dt}{4}\]
\[Now, \int\frac{\left( 1 + \frac{1}{x^4} \right)^{- \frac{3}{4}}}{x^5}\text{ dx }\]
\[ = - \frac{1}{4} \int t^{- \frac{3}{4}} \text{ dt }\]
\[ = - \frac{1}{4} \left[ \frac{t^{- \frac{3}{4} + 1}}{- \frac{3}{4} + 1} \right] + C\]
\[ = - t^\frac{1}{4} + C\]
\[ = - \left( 1 + \frac{1}{x^4} \right)^\frac{1}{4} + C\]