how to integrate 1/(1+x) ?

Question

As title says

Answer 1

Do it by substitution:

Let u = 1+x, so du/dx = 1 and du = dx, so the integral becomes

Integral [1/u du], which is ln(u) + C

Plugging back in for u, we get ln(1+x) + C

Answer 2

Integral of 1/(1+x) dx
= ln(1+x) + c

Answer 3

let u=1+x
du=d(1+x)=dx
Sdx/1+x
=Sdu/u
=lnu + c ; {where c=constant of integration}
=ln(1+x) +c ; {after substituting the value of u}

Answer 4

∫ 1 / (1 + x) dx = ln (1 + x) + C

Answer 5

let 1+x=u then the integral becomes du/u=lnu .u is positive number.

Answer 6

u = 1+x
du = dx
int du/u= ln lul+c =ln l1+xl+c

Answer 7

To find the antiderivative, you must know that whenever there is 1/(x), you use natural log.
So, antiderivative of this is :
Ln(1 + x) + C

Answer 8

u=1+x
du=0+dx
du=dx

integrate: du/u
u^-1du=-ù^-2/2
=-0.5*(1/(1+x)^2))

Answer 9

I remember way back when I was in school. I did my own homework. Do you know what? I still remember how to solve these problems. Please do your own work. The benefit will be yours.