$\int \frac{dx}{x^3\sqrt{(1 + x^4)}} =$

Find $\int \frac{dx}{x^3\sqrt{(1 + x^4)}}$

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Looking at the integral with $x^3$ in the denominator and $x^4$ inside the square root, use the substitution $t = \frac{1}{x^2}$.

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If $t = \frac{1}{x^2} = x^{-2}$

$\frac{dt}{dx} = -2x^{-3}$

$\frac{dt}{dx} = -\frac{2}{x^3}$

$dt = -\frac{2}{x^3}dx$

$\frac{dx}{x^3} = -\frac{1}{2}dt$

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Since $t = \frac{1}{x^2}$, then $x^2 = \frac{1}{t}$ and $x^4 = \frac{1}{t^2}$

$\sqrt{1 + x^4} = \sqrt{1 + \frac{1}{t^2}}$

$= \sqrt{\frac{t^2 + 1}{t^2}}$

$= \frac{\sqrt{t^2 + 1}}{t}$

(assuming $x > 0$, so $t > 0$)

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The integral becomes:

$\int \frac{1}{\frac{\sqrt{t^2 + 1}}{t}} \cdot \left(-\frac{1}{2}\right)dt$

$= -\frac{1}{2}\int \frac{t}{\sqrt{t^2 + 1}} dt$

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Let $v = t^2 + 1$

$dv = 2t \, dt$

$t \, dt = \frac{1}{2}dv$

The integral becomes:

$-\frac{1}{2}\int \frac{1}{\sqrt{v}} \cdot \frac{1}{2}dv$

$= -\frac{1}{4}\int v^{-1/2}dv$

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$-\frac{1}{4} \cdot \frac{v^{1/2}}{1/2} + C$

$= -\frac{1}{4} \cdot 2\sqrt{v} + C$

$= -\frac{1}{2}\sqrt{v} + C$

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Replace $v = t^2 + 1$:

$-\frac{1}{2}\sqrt{t^2 + 1} + C$

Replace $t = \frac{1}{x^2}$:

$-\frac{1}{2}\sqrt{\frac{1}{x^4} + 1} + C$

$= -\frac{1}{2}\sqrt{\frac{1 + x^4}{x^4}} + C$

$= -\frac{1}{2} \cdot \frac{\sqrt{1 + x^4}}{x^2} + C$

$= -\frac{\sqrt{1 + x^4}}{2x^2} + C$