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The substitution $x^2=\sec\theta$ doesn't seem to lead anywhere. I know the key is to manipulate it into a form $$\displaystyle \int \dfrac{\text{d}\left(x^a\pm\frac{1}{x^a}\right)}{f\left(x^a\pm\frac{1}{x^a}\right)}$$

But I fail to do so.

Riz222
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    If I'm not mistaken, this trick works when you have $x^4\color{red}{+}1$ in denominator. You can also use something called Ostrogradsky's method to reduce it to the integral of $1/(x^4-1)$. – bjorn93 Jan 06 '20 at 14:27
  • @bjorn93 this trick works when the denominator is of the form $x^4+kx^2+1$. – Integreek Dec 07 '24 at 12:46
  • Related: https://math.stackexchange.com/a/5000644/1361661 – Integreek Dec 07 '24 at 13:56

3 Answers3

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$$\int\dfrac{dx}{(x^4-1)^2}=\int\dfrac{x^3}{(x^4-1)^2}\cdot\dfrac1{x^3}dx$$

$$=\dfrac1{x^3}\cdot\int \dfrac{x^3}{(x^4-1)^2} dx+\dfrac34\int\dfrac{dx}{x^4(x^4-1)}$$

We can use $$\int\dfrac{dx}{x^4(x^4-1)}=\int\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}dx$$

and $\displaystyle\int\dfrac2{x^4-1}=\int\dfrac{x^2+1-(x^2-1)}{x^4-1}=?$

lab bhattacharjee
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$$\begin{align*} & \int \frac{dx}{\left(x^4-1\right)^2} \\ &= \frac1{16} \int \frac{(1+y)^3}{y^2} \frac{dy}{\sqrt{1-y^2}} & x^2=\frac{1-y}{1+y} \\ &= \frac1{16} \int \frac{(1+\sin z)^3}{\sin^2z} \, dz & y=\sin z \\ &= \frac1{16} \int \left(\csc^2z + 3\csc z + 3 + \sin z\right) \, dz \end{align*}$$

assuming the last step doesn't count as partial fraction expansion.

user170231
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Consider the function

$$\mathcal I(a)=\int_0^x\frac{\mathrm dt}{t^4-a^4}\tag{1}$$

Since $(t^4-a^4)=(t^2+a^2)(t^2-a^2)$, another expression for $\mathcal I(a)$ is

$$\mathcal I(a)=\frac1{2a^3}\tan^{-1}\left(\frac{x}{a}\right)+\frac1{2a^3}\tanh^{-1}\left(\left(\frac{x}a\right)^{\text{sgn}(a^2-x^2)}\right)\tag{2}$$

Equate the expressions for $\mathcal I'(a)$ obtained from $(1)$ and $(2)$, then convert the definite integral into an indefinite one by using the $2^\text{nd}$ part of the FTC, and finally put $a=1$ to obtain the desired result.

Integreek
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