Why can't we integrate $\int\frac{1}{1+x^2}$ to $\frac{\ln(1+x^2)}{2x}$? $\int\frac{1}{1+2x}$ is integrated to $\frac{\ln(1+2x)}{2}$. As we are multiplying and dividing by the derivative of the function inside, why is it not possible to do the same with non-linear polynomials. For example, why can't we do the following:
$\int\frac{1}{1+x^2}$ = $\frac{\ln(1+x^2)}{2x}$
 A: Since Integration should be the inverse of differentiation and vice-versa .
A: The formula is $\int \frac{1}{x}dx=\ln(x)$. In other words, for the result you have used to be true, the denominator of the integrand must be a linear term.
See, what is actually happening in the case of $\int\frac{1}{1+2x}dx$ is as follows:
$$$$
First substitute $u=1+2x$. Thus $du=2dx\Rightarrow dx=\frac{1}{2}du$. Therefore the original integral is transformed into the following:
$$\int\frac{1}{1+2x}dx=\int \frac{1}{u}\left(\frac{1}{2}du\right)=\frac{1}{2}\int \frac{1}{u}du=\frac{\ln(u)}{2}+C=\frac{\ln(1+2x)}{2}+C$$$$$$
In case of $\int\frac{1}{1+x^2}dx$ you can see that the denominator of the integrand is clearly not a linear term. Thus, you cannot directly use the formula $\int \frac{1}{x}dx=\ln(x)$ and equate the integral with $\frac{\ln(1+x^2)}{2x}$.
I would suggest that you learn Integration by Substituiton first so as to understand what is happening.
A: Because there is no rule saying $$\int\frac{1}{f(x)}dx = \frac{\ln(f(x))}{f'(x)}$$
This is in fact only true in special cases.
In your particular case, it is not true because the derivative of $\frac{\ln(1+x^2)}{2x}$ is not $\frac{1}{1+x^2}$, but rather $$\frac{\frac{2x}{1+x^2}\cdot 2x- 2\cdot \ln(1+x^2)}{4x^2}$$
which simplifies to $$\frac{1}{1+x^2}-\frac{\ln(1+x^2)}{2x^2}$$
which is certainly not what you would want it to be.
A: Because
$\frac{d}{dx}\frac{\ln(1+x^2)}{2x}=
\frac{1}{x^2+1}-\frac{\ln(x^2+1)}{2 x^2}$
And as you know:
$\frac{1}{x^2+1}-\frac{\ln(x^2+1)}{2 x^2}\neq\frac{1}{x^2+1}$
