Integral von $\frac{1}{1 - x^{2}}$

Finden Sie $\int \frac{1}{1 - x^{2}}\, dx$.

Perform partial fraction decomposition (steps can be seen »):

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

Integrate term by term:

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

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=\frac{1}{2}$ and $f{\left(x \right)} = \frac{1}{x + 1}$:

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

Let $u=x + 1$.

Then $du=\left(x + 1\right)^{\prime }dx = 1 dx$ (steps can be seen »), and we have that $dx = du$.

Deshalb,

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

The integral of $\frac{1}{u}$ is $\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$:

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

Recall that $u=x + 1$:

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

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=\frac{1}{2}$ and $f{\left(x \right)} = \frac{1}{x - 1}$:

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

Let $u=x - 1$.

Then $du=\left(x - 1\right)^{\prime }dx = 1 dx$ (steps can be seen »), and we have that $dx = du$.

The integral can be rewritten as

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

The integral of $\frac{1}{u}$ is $\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$:

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

Recall that $u=x - 1$:

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

Deshalb,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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