Integrate #x/(x^2+2)^2# ?

1 Answer
Apr 4, 2017

#-1/(2(x^2+2))+C#

Explanation:

We use the substitution rule. We define a new variable #u# as #u=x^2+2#. This implies that #du=2x\ dx#, or #dx=(du)/(2x)#.

We substitute these values in our original problem #int\ x/(x^2+2)^2\ dx# to get #int\ (x\ du)/(2xu^2)#. Canceling out the #x#'s, we get #int\ (du)/(2u^2)#.

This becomes a simple matter of integrating #1/2u^-2# with respect to #u#. Using the constant rule and power rule in integration, we obtain #-1*1/2u^-1+C=-1/(2u)+C#.

However, we want our answer in terms of #x#. Remember that we first defined #u# as #x^2+2#. We just need to substitute this back to get #-1/(2(x^2+2))+C#.