Question

$\int \frac{dx}{x}=\text{\_\_\_\_\_\_\_\_\_}$

1. $x+k$
2. $\frac{1}{x^2}+k$
3. $-\frac{1}{x^2}+k$
4. $logx+k$

Answer: $logx+k$

Solution :-

And we know that integration is an inverse process of Differentiation.
Now Integrate eq(1), we get $\int \frac{d\left(logx+k\right)}{dx}\cdot dx=\int \frac{1}{x}\cdot dx$

Hence, $logx+k=\int \frac{1}{x}\cdot dx\left(\because \int \frac{d\left(f\left(x\right)\right)}{dx}\cdot dx=f\left(x\right)\right)$

So, correct answer is option-4, $\to logx+k$