$$\int\frac{1}{y}\frac{1}{\ln y}dy$$
$$\int\frac{1}{y}\frac{1}{\ln^2y}dy$$
I know that i should use some substitution, but i don't understand how.Is there any other way than substitution?
I've tried to understand the integral-calculator.com output, but no luck.
Can anyone explain to me how to solve these integrals?
On
Using substitution with $u = \ln(y)$ should do the trick. With the chosen $u$ we have that $du = \frac{1}{y}dy$, then we can rewrite the integral as:
$$ \int \frac{dy}{y\ln(y)} = \int \frac{du}{u} $$
Which is easy to integrate. With the other integral, we can use the same substitution
$$ \int \frac{dy}{y\ln^2(y)} = \int \frac{du}{u^2} $$
We have that $(\ln(y))'=\dfrac{1}{y}$.
It is well known that for every differentiable function $f$ and $n\neq -1$, $$\displaystyle\int f'(x)f(x)^n=\dfrac{f(x)^{n+1}}{n+1}+C$$
Therefore, if $k\in\mathbb{Z}$ and $k\neq 1$, we will have $$\displaystyle\int \dfrac{1}{y\ln(y)^k}=\dfrac{\ln(y)^{1-k}}{1-k}+C$$
If $k=1$, $$\dfrac{1}{y\ln(y)}=\dfrac{1/y}{\ln(y)}$$
As the numerator is the derivative of the denominator, primitives will be of the form $\ln(|\ln(y)|)+C$.