how would I go about integrating:
[x^2 / (1+x^2)] dx
And if you didn't know that, you could use trigonometric substitution.
$\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = \int{1\,dx} - \int{\frac{1}{1 + x^2}\,dx}$
$\displaystyle = x - \int{\frac{1}{1 + x^2}\,dx}$.
To find $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$
Let $\displaystyle x = \tan{\theta}$ so that $\displaystyle dx = \sec^2{\theta}\,d\theta$.
Also note that $\displaystyle \theta = \arctan{x}$.
So the integral becomes
$\displaystyle \int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{1}{\sec^2{\theta}}\,\sec^2{\theta}\,d\t heta}$
$\displaystyle = \int{1\,d\theta}$
$\displaystyle = \theta + C$
$\displaystyle = \arctan{x} + C$.
So $\displaystyle \int{1 - \frac{1}{1 + x^2}\,dx} = x - \arctan{x} + C$.
Watch carefully:
$\displaystyle \int{\frac{1}{x}\,dx} = \ln{|x|} + C$ by definition.
We can also see that
$\displaystyle \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\ln{|ax + b|}$.
This comes from a $\displaystyle u$ substitution.
Let $\displaystyle u = ax + b$ so that $\displaystyle \frac{du}{dx} = a$.
So $\displaystyle \int{\frac{1}{ax + b}\,dx} = \frac{1}{a}\int{\frac{a}{ax + b}\,dx}$
$\displaystyle = \frac{1}{a}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$
$\displaystyle = \frac{1}{a}\int{\frac{1}{u}\,du}$
$\displaystyle = \frac{1}{a}\ln{|u|} + C$
$\displaystyle = \frac{1}{a}\ln{|ax + b|} + C$.
BUT you are trying to find $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$.
We DO NOT have a formula for that yet because it is not of the form $\displaystyle \frac{1}{ax + b}$. But you could TRY a $\displaystyle u$ substitution.
Watch what happens.
Let $\displaystyle u = 1 + x^2$.
We can see that $\displaystyle \frac{du}{dx} = 2x$.
Can we change $\displaystyle \int{\frac{1}{1 + x^2}\,dx}$ into a function that is of the form $\displaystyle \int{\frac{2x}{1 + x^2}\,dx}$? NO, we can not.
So the rule for $\displaystyle \ln$ DOES NOT WORK in this case.
Now see my post regarding trigonometric substitution to show you the method that DOES work.
Alternatively, let's try differentiating $\displaystyle \arctan{x}$ and seeing if we get $\displaystyle \frac{1}{1 + x^2}$. If we do, then $\displaystyle \int{\frac{1}{1 + x^2}\,dx} = \arctan{x}$.
$\displaystyle y = \arctan{x}$
$\displaystyle \tan{y} = x$
$\displaystyle \frac{d}{dx}(\tan{y}) = \frac{d}{dx}(x)$
$\displaystyle \frac{dy}{dx}\,\frac{d}{dy}(\tan{y}) = 1$
$\displaystyle \sec^2{y}\,\frac{dy}{dx} = 1$
$\displaystyle \frac{dy}{dx} = \cos^2{y}$.
But we know that $\displaystyle y = \arctan{x}$ and we also know that $\displaystyle \cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}$
So $\displaystyle \frac{dy}{dx} = \left(\frac{1}{\sqrt{1 + \tan^2{y}}}\right)^2$
$\displaystyle = \frac{1}{1 + \tan^2{y}}$
$\displaystyle = \frac{1}{1 + (\tan{\arctan{x}})^2}$
$\displaystyle = \frac{1}{1 + x^2}$.
So $\displaystyle \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}$.
Therefore $\displaystyle \int{\frac{1}{1 + x^2}\,dx} = \arctan{x} + C$.
No more arguments please!