what is the anti derivative of x/(x^2+1) and why?

Question

hartnn · Answer

try 
u= x^2+1
du= ..?

anonymous · Answer

du= 2x

hartnn · Answer

only 2x ?
or 
du = 2x dx ?

anonymous · Answer

yes 2x dx

hartnn · Answer

so,
x dx = du/2
so whats your new integral in terms of 'u' only ?

anonymous · Answer

2x/2 dx

anonymous · Answer

or 2x dx/2..

hartnn · Answer

what ?

hartnn · Answer

sorry i ddn't get what doubt you have ..

anonymous · Answer

so in respect to u it would be 2x/2 then

hartnn · Answer

i am not sure on how you get 2x/2
let me show you steps

hartnn · Answer

u = x^2+1 
du= 2x dx 
xdx = du/2  which is your numerator,
denominator = x^2+1 = u 
so your new integral is 
$\int \dfrac{du}{2u}$
got this ?

anonymous · Answer

ok yes

hartnn · Answer

can you integrate that ?

hartnn · Answer

$\int (1/x)dx = \ln |x|+c

 $

anonymous · Answer

the answer on the example is $$\lim_{b \rightarrow \infty} [\frac{ 1 }{ 2}\ln(x ^{2}+1)]_{1}^{b}$$

hartnn · Answer

ok

hartnn · Answer

can you integrate 
$\int \dfrac{du}{2u}$
?

anonymous · Answer

$$\int\limits_{?}^{?} \frac{ 1 }{ 2 }*\frac{ 1 }{ x^2+1 }= 1/2 \ln(x^2+1)$$

hartnn · Answer

$\int du/2u  = 1/2 \ln |u|+c$
but u = x^2+1
so, 
$= 1/2 \ln |x^2+1| +c$

hartnn · Answer

and if you are given limits from 1 to infinity
you get 
$\lim_{b \rightarrow \infty} [\frac{ 1 }{ 2}\ln(x ^{2}+1)]_{1}^{b}$

anonymous · Answer

thank you, btw what is the rule that you used earlier called? the du/u stuff?

hartnn · Answer

its not called anything, its just a formula...

anonymous · Answer

gotcha