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## Main Question or Discussion Point

**Calculus help plz!!! another one**

Can someone plz show me how to show that this problem below is CORRECT? Thanks a lot.

Integral of dx/(1-x^2) = x/(1-x)

- Thread starter gigi9
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Can someone plz show me how to show that this problem below is CORRECT? Thanks a lot.

Integral of dx/(1-x^2) = x/(1-x)

- #2

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Use the substitution x=sinu.

- #3

Soroban

Hello, gigi9!

The answer is not correct . . .

We can use Partial Fractions: 1/(1 - x^2) = A/(1 - x) + B/(1 + x)Originally posted by gigi9

Can someone plz show me how to show that this problem below is CORRECT?

Integral of dx/(1-x^2) = x/(1-x)

and find that: A = 1/2, B = -1/2.

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

- #4

Hurkyl

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The easiest way to check if an integral is correct is to invert the operation and differentiate.Can someone plz show me how to show that this problem below is CORRECT? Thanks a lot.

Integral of dx/(1-x^2) = x/(1-x)

Via the fundamental theorem of calculus, if

∫ 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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