Question

the primitive of ln(x)/1+x^2 ??

Answer 1

x*ln(x) - x + (x^3)/3
Integrate ln(x) and x^2 separately.
\[\int\limits_{}^{} \ln(x) + x^2 dx = \int\limits_{ }^{ } \ln (x) dx + \int\limits_{ }^{ } x^2 dx\]
The second part there is easy:(x^3)/3. Put that off to the side.
The first one is a bit more complicated. You must use a method called "integration by parts." Here's the equation you use (sorry I don't know the full proof)
\[\int\limits_{}^{}udv = uv - \int\limits_{}^{} vdu\]
Consider ln(x) to be v and dx to be dv, thus:
u = ln(x) v = x
du = 1/x dx dv = dx
Plugging all of those values into the equation, you get:
\[xln(x) - \int\limits_{}^{} dx \rightarrow xln(x) -x\]
And then just add the other part you shoved off to the side, (x^3)/3.