Question

$\int \frac{lnx-1}{(lnx)^2}dx$

Answer 1

$\frac{x}{\ln x} + C$

Answer 2

To evaluate the integral $\int \frac{\ln x - 1}{(\ln x)^2} dx$, we can use a substitution method.

Let $t = \ln x$.
From this substitution, we can express $x$ in terms of $t$:
$x = e^t$.

Now, differentiate $x = e^t$ with respect to $t$ to find $dx$:
$dx = e^t dt$.

Substitute $t$ and $dx$ into the integral: $\int \frac{\ln x - 1}{(\ln x)^2} dx = \int \frac{t - 1}{t^2} e^t dt$

Rearrange the terms inside the integral: $\int \left(\frac{t}{t^2} - \frac{1}{t^2}\right) e^t dt = \int \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt$

This integral is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
In our case, let $f(t) = \frac{1}{t}$.
Then, the derivative of $f(t)$ with respect to $t$ is $f'(t) = -\frac{1}{t^2}$.

So, the integral matches the form $\int e^t (f(t) + f'(t)) dt$.
Applying the formula, the integral evaluates to: $e^t f(t) + C = e^t \cdot \frac{1}{t} + C$

Finally, substitute back $t = \ln x$ into the result: $e^{\ln x} \cdot \frac{1}{\ln x} + C$ Since $e^{\ln x} = x$, the expression simplifies to: $x \cdot \frac{1}{\ln x} + C = \frac{x}{\ln x} + C$

To verify the result, differentiate $\frac{x}{\ln x}$ with respect to $x$ using the quotient rule:
Let $u = x$ and $v = \ln x$. Then $u' = 1$ and $v' = \frac{1}{x}$. $\frac{d}{dx}\left(\frac{x}{\ln x}\right) = \frac{u'v - uv'}{v^2} = \frac{1 \cdot \ln x - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}$ This matches the original integrand, confirming the correctness of the solution.