Evaluate the integral ? e^1/(x+1)/(x+1)^2 dx

[venialove]

Evaluate the integral ? e^1/(x+1)/(x+1)^2 dx

My work
x+1(0)-e^1(x+1)(2)/(x+1)^2
= e^-x/(x+1)/(x+1)^2+c

 

[stapel]

e^1/(x+1)/(x+1)^2

I'm sorry, but the formatting of your integrand makes no sense, that I'm aware of. You can't have three fraction bars in a fraction (at least not without plenty of parentheses, etc, to clarify which bits actually go together). :shock:

Please review the articles in the links you noted when you read the "Read Before Posting" article that you had read before posting, and kindly reply with clarirification in standard web-safe formatting. Or else please study the LaTeX articles in the "Forum Help" pull-down menu at the very top of every forum page, and repost using that.

Thank you!

Eliz.

 

[skeeter]

you probably meant the integrand to be ...

[e[sup:19a961n7]1/(x+1)[/sup:19a961n7]]/(x+1)[sup:19a961n7]2[/sup:19a961n7]

use the method of substitution ...

let u = 1/(x+1) ... du = -1/(x+1)[sup:19a961n7]2[/sup:19a961n7] dx