Question

FACT: d/dx((1-x)/1+x))= (-2)/((1+x)^2).
Then the integral of (-2/((1+x)^2)))dx = ((1-x)/(1+x)) + c

EVALUATE : the integral of ((-2/((1+x)^2)) dx by letting u=1+x and du=dx
Do you get the same answer as above? EXPLAIN.

Answer 1

WELL.......DX=476 AND EX=209

Answer 2

that doesn't even make sense

Answer 3

the answers may or may not look the same
but the answer should be equivalent

Answer 4

I have to the point -2/u^2 dx = 2/(1+x) ?

Answer 5

I guess with a +c so 2/(1+x) +c but I am not sure how this is equivelant

Answer 6

\[\frac{1-x}{1+x}+C=\frac{1+x-2x}{1+x}+C=\frac{1+x}{1+x}-\frac{2x}{1+x}+C \\ 1-\frac{2x}{1+x}+C \\ =-\frac{2x}{1+x}+K_1 \text{ since } 1+C \text{ is still a constant } \\ =\frac{-2(x+1)+2}{1+x}+K_1 =-2+\frac{2}{1+x}+K_1 \\ =\frac{2}{1+x}+K \text{ since } -2+K_1 \text{ is still a constant }\]

Answer 7

thought you could have just divided (1-x) by (1+x) to get the same result there

Answer 8

though*