$$$\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}}$$$ 的積分

求$$$\int \frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}}\, dx$$$。

進行部分分式分解（步驟可見 »）:

$${\color{red}{\int{\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}} d x}}} = {\color{red}{\int{\left(- \frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} + \frac{1}{4 \left(2 x + \sqrt{2}\right)^{2}} + \frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} + \frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}}\right)d x}}}$$

逐項積分：

$${\color{red}{\int{\left(- \frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} + \frac{1}{4 \left(2 x + \sqrt{2}\right)^{2}} + \frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} + \frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}}\right)d x}}} = {\color{red}{\left(\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \int{\frac{1}{4 \left(2 x + \sqrt{2}\right)^{2}} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(x \right)} = \frac{1}{\left(2 x + \sqrt{2}\right)^{2}}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + {\color{red}{\int{\frac{1}{4 \left(2 x + \sqrt{2}\right)^{2}} d x}}} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + {\color{red}{\left(\frac{\int{\frac{1}{\left(2 x + \sqrt{2}\right)^{2}} d x}}{4}\right)}}$$

令 $$$u=2 x + \sqrt{2}$$$。

則 $$$du=\left(2 x + \sqrt{2}\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

該積分可改寫為

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\int{\frac{1}{\left(2 x + \sqrt{2}\right)^{2}} d x}}}}{4} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\int{\frac{1}{2 u^{2}} d u}}}}{4}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\int{\frac{1}{2 u^{2}} d u}}}}{4} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u^{2}} d u}}{2}\right)}}}{4}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{8}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\int{u^{-2} d u}}}}{8}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{8}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{8}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{8}$$

回顧一下 $$$u=2 x + \sqrt{2}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} - \frac{{\color{red}{u}}^{-1}}{8} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x} - \frac{{\color{red}{\left(2 x + \sqrt{2}\right)}}^{-1}}{8}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(x \right)} = \frac{1}{\left(2 x - \sqrt{2}\right)^{2}}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + {\color{red}{\int{\frac{1}{4 \left(2 x - \sqrt{2}\right)^{2}} d x}}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{\left(2 x - \sqrt{2}\right)^{2}} d x}}{4}\right)}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}$$

令 $$$u=2 x - \sqrt{2}$$$。

則 $$$du=\left(2 x - \sqrt{2}\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

所以，

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{\left(2 x - \sqrt{2}\right)^{2}} d x}}}}{4} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{2 u^{2}} d u}}}}{4} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{2 u^{2}} d u}}}}{4} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u^{2}} d u}}{2}\right)}}}{4} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\int{u^{-2} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}=\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}$$

回顧一下 $$$u=2 x - \sqrt{2}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} - \frac{{\color{red}{u}}^{-1}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x} - \frac{{\color{red}{\left(2 x - \sqrt{2}\right)}}^{-1}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{\sqrt{2}}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{2 x + \sqrt{2}}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - {\color{red}{\int{\frac{\sqrt{2}}{8 \left(2 x + \sqrt{2}\right)} d x}}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{2 x + \sqrt{2}} d x}}{8}\right)}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

令 $$$u=2 x + \sqrt{2}$$$。

則 $$$du=\left(2 x + \sqrt{2}\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

該積分變為

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 x + \sqrt{2}} d x}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 u} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 u} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

回顧一下 $$$u=2 x + \sqrt{2}$$$：

$$- \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} + \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(2 x + \sqrt{2}\right)}}}\right| \right)}}{16} + \int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{\sqrt{2}}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{2 x - \sqrt{2}}$$$：

$$- \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + {\color{red}{\int{\frac{\sqrt{2}}{8 \left(2 x - \sqrt{2}\right)} d x}}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{2 x - \sqrt{2}} d x}}{8}\right)}} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

令 $$$u=2 x - \sqrt{2}$$$。

則 $$$du=\left(2 x - \sqrt{2}\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

因此，

$$- \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 x - \sqrt{2}} d x}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 u} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$- \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{2 u} d u}}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{8} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$- \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

回顧一下 $$$u=2 x - \sqrt{2}$$$：

$$- \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)} = - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(2 x - \sqrt{2}\right)}}}\right| \right)}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

因此，

$$\int{\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}} d x} = \frac{\sqrt{2} \ln{\left(\left|{2 x - \sqrt{2}}\right| \right)}}{16} - \frac{\sqrt{2} \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}}{16} - \frac{1}{8 \left(2 x + \sqrt{2}\right)} - \frac{1}{8 \left(2 x - \sqrt{2}\right)}$$

化簡：

$$\int{\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}} d x} = \frac{- 8 x + \sqrt{2} \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right) \left(\ln{\left(\left|{2 x - \sqrt{2}}\right| \right)} - \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}\right)}{16 \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)}$$

加上積分常數：

$$\int{\frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}} d x} = \frac{- 8 x + \sqrt{2} \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right) \left(\ln{\left(\left|{2 x - \sqrt{2}}\right| \right)} - \ln{\left(\left|{2 x + \sqrt{2}}\right| \right)}\right)}{16 \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)}+C$$

$$$\int \frac{x^{2}}{\left(2 x^{2} - 1\right)^{2}}\, dx = \frac{- 8 x + \sqrt{2} \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right) \left(\ln\left(\left|{2 x - \sqrt{2}}\right|\right) - \ln\left(\left|{2 x + \sqrt{2}}\right|\right)\right)}{16 \left(2 x - \sqrt{2}\right) \left(2 x + \sqrt{2}\right)} + C$$$A