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

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

#### 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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