how do you derive the integral of 1/1+x^n?..what are the steps?

Answer 1

Here's the problem; the integral *heavily* depends on what n is.

Case 1: n = 0.

Integral (1/(1 + 0) dx) = Integral (1)dx = x + C

Case 2: n = 1:

Integral (1 / (1 + x)) dx = ln |1 + x| + C

Case 3: n = 2:

Integral (1 / (1 + x^2)) dx = arctan(x) + C

Case 4: n = 3:

Integral (1 / (1 + x^3)) dx, and this may not even be expressable in terms of elementary functions.

How about if n = -1? Then

Integral (1 / (1 + x^(-1))dx = Integral (x / (x + 1)) dx
= Integral (1/(x + 1) + 1) dx
= ln|x + 1| + x + C

Let's see what happens if n = -2:

Integral (1 / (1 + x^(-2)) ) dx =
Integral ( (x^2) / (x^2 + 1) ) dx =
Integral ( [1 / (x^2 + 1)] + 1 ) dx =

arctan(x) + x + C

Then there are the fractional cases, like n = 1/2, 1/3, etc... which, also, may not be expressable as elementary functions.

The bottom line is that, without great difficulty, there is no way to find a general formula for the integral of (1 / (1 + x^n)).

Answer 2

For this question the integral of 1/(1+x^n) is:
it can be split into sum of the two integrals as per the
integral of [f(x)+g(x)]dx= integral of f(x)dx+integral of g(x)dx
so this can be split as integral of 1/1dx+integral of 1/x^ndx
now integral 1/1dx=x and integral 1/x^n dx = -[ 1/ (n-1)x^(n-1)]+C
integral 1/1+x^n = x-[1/(n-1)x^(n-1)]+C.(where n is not equal to 1)
IF n=1 then 1/x^n = log x + C (since 1/x^1=1/x)
integral 1/1+x^1= x+log x + C (n=1)

Answer 3

There is a general expression for your integral:

x * Hipergeometric2F1(1/n, 1, 1+ 1/n, - x^n)

but that is to say that the integral cannot be expressed by familiar functions.