Why is ∫x/(1+x^2)dx = 1/2ln(1+x^2)?

[5P@N]

why is ∫x/(1+x^2) dx = 1/2ln(1+x^2)?

If you could just show me a couple of relevant integration rules, that would be great. I'm having trouble figuring this one out.

 

[blue_leaf77]

Ever heard of substitution method http://www.mathportal.org/calculus/integration-techniques/integration-by-substitution.php?

 

[Buzz Bloom]

Hi 5P:

Let u = (1+x^2). Then, what is du? With this substitution, what form does the integral then take?

Hope this helps.

Regards,
Buzz

 

[fresh_42]

It's the same substitution as before: $u(x) = 1 + x^2$ and therefore $u' = du / dx = 2x$ which gives you $x dx = \frac{1}{2} du$ and then $1/u$ to integrate on $u$, i.e. $ln|u|$.

 

[Mark44]

If you could just show me a couple of relevant integration rules

You should already have seen the substitution method, which is one of the first techniques that are presented. Also, it's also one that you should try first when you have an integration problem. This technique might not be useful in some cases, but it's reasonably simple, so if it doesn't work, you haven't wasted much time.