# Integration Problem

#### natHenderson

##### New member

So far I've tried u-substitution and set the u = e^x - 1. This gave me the integral of 1/(1+u^2)du which is arctan(u) or arctan(e^x-1), then i set the pike from 0 to ln(2), but that just gave me arctan(3) - arctan(2) which isn't one of the answers, does anybody know what I did wrong?

#### Subhotosh Khan

##### Super Moderator
Staff member
What did you get when you set the limits? - show detailed work. Your integration is correct. We would not know "what you did wrong" - unless you show us your work regarding the limits.

#### topsquark

##### Full Member
You got the limits wrong.
$$\displaystyle tan^{-1} \left ( e^x - 1 \right )_{x = ln(2)} = tan^{-1} ( 2 - 1) = tan^{-1}(1) = \dfrac{ \pi }{4}$$

Your lower limit is off, too.

-Dan

#### natHenderson

##### New member
Oh ok, thanks Dan!