# Thread: A couple of indefinite integrals

1. ## A couple of indefinite integrals

I've skimmed a few topics in the Calculus forum, but they haven't turned up answers, and I can't really read anything what with all the [tex] notation. So, maybe this way will work.

In some differential equations I've been working on, I found integrals like 1/(exp(x)+1) and 1/(x+x^2), which I apparently forgot how to solve since high school. I racked my brain about substitutions, but I'm only good at getting stuff to cancel out, and without nice numerators, I'm stumped.

Any help is appreciated (especially regarding how to read what you guys respond with).

2. Originally Posted by primasapere
I've skimmed a few topics in the Calculus forum, but they haven't turned up answers, and I can't really read anything what with all the [tex] notation. So, maybe this way will work.

In some differential equations I've been working on, I found integrals like 1/(exp(x)+1) and 1/(x+x^2), which I apparently forgot how to solve since high school. I racked my brain about substitutions, but I'm only good at getting stuff to cancel out, and without nice numerators, I'm stumped.

Any help is appreciated (especially regarding how to read what you guys respond with).
For the second partial fractions should suffice:

1/(x+x^2)=1/x - 1/(x+1)

and for the first try substituting u=exp(x), which should then give you the
second integral.

RonL

3. Don't make double posts, if you don't know where to put your
question put it here. You are just making extra work for the
helpers time by making double posts.

Your other post has been deleted.

RonL

4. That partial fractions method is genius, but the exp(x) substitution didn't seem to work. u = exp(x) gives dx = du/exp(x), then my integral is in terms of u and x.

Here's another tough integral, maybe only because it's late: x^2/(1-kx), with k being an arbitrary constant. Can't sub u = 1-kx, or I get two terms again. Can't sub u = x^2, because it doesn't simplify anything.

Also, if there's any kind of website anyone knows about with some good pointers on solution methods for integration like this, that might save time for both of us.

5. That's a top heavy fraction so I would do algebraic division first to break it down.

6. Originally Posted by primasapere
That partial fractions method is genius, but the exp(x) substitution didn't seem to work. u = exp(x) gives dx = du/exp(x), then my integral is in terms of u and x.
Code:
int 1/(exp(x)+1) dx,
let u=exp(x), then du/dx=u, dx=(1/u)du, so:

Code:
int 1/(exp(x)+1) dx = int 1/(u+1) 1/u du = int 1/(u^2+u) du
RonL