$\int x^{- 11} {(1 + x^{2})}^{\frac{1}{2}} d x is equal to$

Moderate

A

$- \frac{1}{6} {(\frac{1}{x^{4}} + 1)}^{\frac{3}{2}} + 10 {(\frac{1}{x^{4}} + 1)}^{\frac{5}{2}} + C$
B

$- \frac{1}{6} {(\frac{1}{x^{4}} + 1)}^{\frac{- 3}{2}} + \frac{1}{10} {(\frac{1}{x^{4}} + 1)}^{\frac{5}{2}} + C$
C

$\frac{1}{6} {(\frac{1}{x^{4}} + 1)}^{\frac{3}{2}} - \frac{1}{10} {(\frac{1}{x^{4}} + 1)}^{\frac{5}{2}} + C$
D

$\frac{1}{6} {(\frac{1}{x^{4}} + 1)}^{\frac{3}{2}} + \frac{1}{10} {(\frac{1}{x^{4}} + 1)}^{\frac{5}{2}} + C$

Solution

$\int x^{- 11} {(1 + x^{4})}^{\frac{1}{2}} d x$
$\begin{aligned} = \int \frac{\sqrt{1 + x^{4}}}{x^{11}} d x \\ = \int \frac{x^{2} \sqrt{1 + \frac{1}{x^{4}}}}{x^{11}} d x \\ = \int \frac{\sqrt{1 + \frac{1}{x^{4}}}}{x^{9}} d x \\ = \int \frac{\sqrt{1 + \frac{1}{x^{4}}}}{x^{4} x^{5}} d x \end{aligned}$
$Put 1 + \frac{1}{x^{4}} = t^{2}$
$\begin{array}{l} \frac{- 4}{x^{5}} d x = 2 t d t \\ \frac{d x}{x^{5}} = \frac{- 2}{4} t d t \\ = \int t (t^{2} - 1) (- \frac{1}{2} t d t) \end{array}$
$\begin{array}{l} = \frac{- 1}{2} \int t^{2} (t^{2} - 1) d t \\ = \frac{- 1}{2} \int [t^{4} - t^{2}] d t \\ = \frac{- 1}{2} [\frac{t^{5}}{5} - \frac{t^{3}}{3}] + C \\ = \frac{1}{6} {[\frac{1}{x^{4}} + 1]}^{\frac{3}{2}} - \frac{1}{10} {(\frac{1}{x^{4}} + 1)}^{\frac{5}{2}} + C \end{array}$

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