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∫ 1 X 2 ( X 4 + 1 ) 3 / 4 D X - Mathematics

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

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

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.9 | Q 71 | Page 59

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