Q:

How to prepare this function for integration

I want to prepare $$f(x)=\frac{x}{1+x^2}$$ for integration, how do i get the $1+x^2$ to the top?
Is $$\frac{x}{1+x^2}$$ the same as $\frac x1 + \frac{x}{x^2}$? If not please explain how I prepare the following function for integration.

A:

Hint
You do not need to have $1+x^2$ in the top. Just notice that the derivative of the denominator is $2x$ that is to say very similar to the numerator. So, try to put it into something looking as $\frac {u'(x)}{u(x)}$

A:

In general $\displaystyle \frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$.
So you can't simply split up the integrand as you did.
However, it is true that $\displaystyle \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$, so you might be confusing this with that.
To evaluate your integral, simply make the substitution $\displaystyle u = 1 + x^2$. Remember that $du = 2xdx$, so $\displaystyle dx = \frac{1}{2x}du$, which will allow you to greatly simplify the integral.

A:

Is $$\frac{x}{1+x^2}$$ the same as $\frac x1 + \frac{x}{x^2}$?

I'm sorry to say this, but if you do not know that this is absolutely not true, then you must learn A LOT more before you can start learning integration.

