Integrate $\frac{x}{1+x^2}$ Integrating $\frac{x}{1+x^2}$ becomes $\ln \sqrt{x^2 + 1}$
Why is this? Is there a formula or a fact that makes this so. Integrating this a lot different than integrating something easy like $X^2$.
 A: Now that the formula is clear, we can use
$$\int\frac{f'(x)}{f(x)}dx=\log |f(x)|+C$$
since $\;x=\frac12(2x)=\frac12(x^2+1)'\;$ , and thus
$$\int\frac x{x^2+1}dx=\frac12\int\frac{(x^2+1)'}{x^2+1}dx=\frac12\log(x^2+1)+C=\log\sqrt{x^2+1}+C$$
A: Hint: Whenever you see a fractional expression like this where the numerator is very nearly the derivative of the denominator, you should suspect that the expression may be the derivative of the log of the denominator.
There will typically be a constant multiplier that you will have to put in (and also divide by it so you don't change the expression in the end) to get it in the exact form.
The rule to remember is $$\left(\log u\right)' = \dfrac{u'}{u}$$
which is exactly equivalent to the antiderivative rule $$\int \dfrac{du}{u}=\log u +C$$
Hint: Also, the radical symbol inside the log is fluff. It's just the same as a multiplier of $\frac12$ outside the log, because of the rule $$\log u^k = k\log u$$
A: I would say the basic rules of integration include general very helpful relationship
$\displaystyle \int \frac{f'(x)}{f(x)}\,dx=\ln |f(x)| \,\,+C$.
$\displaystyle\Rightarrow \int\frac{x}{x^2+1}\,dx=\frac{1}{2}\int\frac{2x}{x^2+1}\,dx=\frac{1}{2}\int\frac{(x^2+1)'}{x^2+1}\,dx=\frac{1}{2}\ln (x^2+1)=\ln \sqrt{x^2+1}+C$
