Can anyone find this integral?

Question

10/ (1+x)^2

Answer 1

I'm going to use INT for the integral sign.

Take out the 10 and get:

10(INT(1+x)^(-2) = 10*-(1+x)^(-1) = -10/(1+x) + C

Answer 2

Think of it as 10Integral(1 + x)^(-2)dx
The power rule works with this:
Integral [f(x)^n]f'(x)dx = f(x)^(n+1)/(n+1) + c
In your problem f(x) = (1 + x), and f'(x) =1 and n = -2
so your integral is 10 (1 + x)^(-1)/(-1) + c or
-10/(1 + x) + c

Answer 3

Int = -10/(1+x)+C