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Evaluate the integral $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$

My Attempt:

Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to

$$ -\int \frac{t^3}{(1+t^4)^{3/4}}\,dt $$

Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to

$$ \begin{align} -\frac{1}{4}\int t^{-3/4}\,dt &= -u^{1/4}+\mathcal{C}\\ &= -\left(1+t^4\right)^{1/4}+\mathcal{C} \end{align} $$

So we arrive at the solution

$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$

Question: Is there any other method for solving this problem?

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The following method feels more systematic to me:

$$\begin{align} \int\frac{\mathrm{d}x}{x^2\left(1+x^4\right)^{3/4}} &=\frac14\int\frac{4x^3\,\mathrm{d}x}{x^5\left(1+x^4\right)^{3/4}}\\ &=\frac14\int\frac{\mathrm{d}t}{t^{5/4}\left(1+t\right)^{3/4}};~~~\small{x^4=t}\\ &=\frac14\int\left(\frac{1+t}{t}\right)^{-3/4}\cdot\frac{1}{t^2}\,\mathrm{d}t\\ &=-\frac14\int u^{-3/4}\,\mathrm{d}u;~~~\small{\frac{1+t}{t}=u}\\ &=-\sqrt[4]{u}+\color{grey}{constant}\\ &=-\sqrt[4]{\frac{1+t}{t}}+\color{grey}{constant}\\ &=-\sqrt[4]{\frac{1+x^4}{x^4}}+\color{grey}{constant}\\ &=-\frac{\sqrt[4]{1+x^4}}{x}+\color{grey}{constant}.\\ \end{align}$$

But the result is of course the same.

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Alternate Solution:

$$ \begin{align} \int\frac{1}{x^2\cdot \left(x^4+1\right)^{3/4}}\,dx &= \int\frac{1}{x^2\cdot x^3\cdot \left(1+x^{-4}\right)^{3/4}}\,dx\\ &= \int\frac{1}{x^{5}\cdot \left(1+x^{-4}\right)^{3/4}}\,dx \end{align} $$

Now Let $(1+x^{-4}) = t$. Then $x^{-5}\,dx = -\frac{1}{4}\,dt$, so we have

$$ -\frac{1}{4}\int \frac{1}{t^{3/4}}dt = -\frac{1}{4}\cdot 4\cdot t^{1/4}+\mathcal{C} = -\left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C} $$

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Let $$\displaystyle I = \int\frac{1}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx = \int \frac{(x^4+1)-x^4}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx$$

So we get $$\displaystyle I = \int\left[\frac{(x^4+1)^{\frac{1}{4}}-x^4\cdot (x^4+1)^{-\frac{3}{4}}}{x^2}\right]dx = \int \left[\frac{x(-x^3)\cdot (1+x^4)^{-\frac{3}{4}}+(x^4+1)^{\frac{1}{4}}}{x^2}\right]dx$$

So we get $$\displaystyle I = -\int \frac{d}{dx}\left[\frac{(1+x^4)^{\frac{1}{4}}}{x}\right]dx = -\frac{(1+x^4)^{\frac{1}{4}}}{x}+\mathcal{C}$$

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