Question

Dx/(1-x)ln^2(1-x) integral

Answer 1

Ответ:

$\int\limits {\frac{dx }{(1-x)ln^{2}(1-x) } } \, =-\int\limits {\frac{d(1-x) }{(1-x)ln^{2}(1-x) } } \, =-\int\limits {\frac{d(ln(1-x)) }{ln^{2}(1-x) } } \,=-\frac{ln^{-2+1}(1-x)}{-2+1} =\frac{1}{ln(1-x)} +C$

Объяснение:

$\frac{1}{ln(1-x)} +C$

Answer 2

To solve the integral $\int \frac{dx}{(1-x)\ln^2(1-x)}$, we can use substitution. Let's make the substitution $u = 1 - x$, so $du = -dx$. The integral becomes:

$$-\int \frac{du}{u\ln^2(u)}$$

Now, let's make a substitution for the denominator. Let $v = \ln(u)$, so $dv = \frac{1}{u}du$. The integral becomes:

$$-\int \frac{dv}{v^2}$$

Integrating $-\frac{1}{v^2}$ with respect to $v$ gives $\frac{1}{v} + C$, where $C$ is the constant of integration. Substituting back $v = \ln(u)$ and $u = 1 - x$, we get the final answer:

$$-\frac{1}{\ln(1-x)} + C$$

So, the integral of $\frac{dx}{(1-x)\ln^2(1-x)}$ is $-\frac{1}{\ln(1-x)} + C$, where $C$ is the constant of integration.