# Is the Integral of dx/(1-x^2) Equal to x/(1-x)?

• gigi9
The correct answer is:&int; dx / (1 - x^2) = 1/2 ln |(1 + x) / (1 - x)| + CIn summary, the correct answer to the problem is 1/2 ln |(1 + x) / (1 - x)| + C.
gigi9

Can someone please show me how to show that this problem below is CORRECT? Thanks a lot.
Integral of dx/(1-x^2) = x/(1-x)

Use the substitution x=sinu.

Hello, gigi9!

The answer is not correct . . .

Originally posted by gigi9
Can someone please show me how to show that this problem below is CORRECT?
Integral of dx/(1-x^2) = x/(1-x)
We can use Partial Fractions: 1/(1 - x^2) = A/(1 - x) + B/(1 + x)
and find that: A = 1/2, B = -1/2.

The answer will be: (1/2) ln|(1 - x)/(1 + x)| + C

Can someone please show me how to show that this problem below is CORRECT? Thanks a lot.
Integral of dx/(1-x^2) = x/(1-x)

The easiest way to check if an integral is correct is to invert the operation and differentiate.

Via the fundamental theorem of calculus, if

&int; dx / (1 - x^2) = x / (1 - x)

then

1 / (1 - x^2) = (d/dx) (x / (1 - x))

So if we actually perform the differentiation, we get:

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

Because this equation is false, the original problem (as you've written it) must be false as well.

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