integrals. ∫ 1/(1-x²) dx!? please help!?

Question

how do i show that ∫ 1/(1-x²) dx = ln sqrt[(1+x) / (1-x)] + C, valid for |x|<1

i know 2/(1-x) = 1/(1+x) + 1/(1-x) should help in some way!

thanks for any help

Answer 1

1/(1-x^2) = 1/(x+1)(x-1) = 1/2 [ 1/(x+1) -1/(x-1) }

Hence integral is 1/2 { ln (x+1) -ln(x-1) } because int of 1/x is lnx


but lnm-ln n = ln ln m/n

& 1/2 ln z= ln z ^1/2

Thus we can simplyfy the answer as

ln sqrt{( 1+x)/(1-x) } + C (the constant of intergration)

Answer 2

∫ 1/(1-x²) dx

in this case, i used partial fractions

1/(1-x²) = A/(1+x) + B/(1-x)

A(1-x) + B(1+x) = A - Ax + B +Bx
= (A+B) - x(A-B)

we now have the equations

A + B = 1
A - B= 0

Solving for A and B,
A = 1/2
B = -1/2

therefore we now have 1/(1-x²) = 1/2*(1+x) - 1/2*(1-x)

solving the integral of 1/(1-x²)

∫ 1/(1-x²) dx = 1/2∫ 1/(1+x) dx - 1/2∫ 1/(1-x) dx

= 1/2 ln|1+x| - 1/2 ln|1-x| + C

= 1/2 ln| (1+x)/(1-x)| + C

= ln sqrt[(1+x)/(1-x)] + C

* C is some constant.

Answer 3

In this case it is best to work backwards. Since you know that:

∫ 1/(1-x²) dx = ln sqrt[(1+x) / (1-x)] + C

Then you know that d/dx[ln sqrt[(1+x) / (1-x)] + C]=1/(1-x²)

Since integration is just the opposite of differentiation. So lets differentiate ln sqrt[(1+x) / (1-x)] + C;

The C just dissappears, and using the rule of logs we can simplify that logarithm to:

1/2[ln(1+x) - ln(1-x)]

Differentiating that in your head or using the chain rule if necessary gets you:

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

And doing some algebraic manipulation [namely putting over the same denomenator and performing the subtraction] gets you:

1/(1-x²)

Which is exactly what you want.

Of course its only valid for |x|<1 since sqrt[(1+x) / (1-x)] becomes complex if |x|>=1

Answer 4

write that and cut into parts!
we know that : (1-x^2)=(1+x)(1-x)

and we can write that:
A/(1+x)+B/(1-x)=1/(1-x^2)
A(1-x)+B(1+x)=1
=> A=B=1/2

∫ [ (1/2)*(1/ (1-x) ) ] dx +∫ [ (1/2)*(1/ (1+x) ) ] dx =

we know that too: ∫ [ 1 / (1-x) ] dx = - ln(1-x)
∫ [ 1 / (1+x) ] dx = ln(1+x)

hence:

(-1/2)*ln(1-x) + (1/2) *ln(1+x)+c = - ln sqrt(1-x) + ln sqrt(1+x) +c
= ln [(1+x)/(1-x)] + c

Answer 5

1 / (1 - x).(1 + x) = A / (1 - x) + B / (1 + x)
1 = A (1 + x) + B (1 - x)
1 = (A - B).x + (A + B)
A - B = 0
A + B = 1
2A = 1
A = 1/2
B = 1/2
I = (1/2) ∫ dx / (1 - x) + (1/2) ∫dx / (1 + x)
I = (- 1/2) log(1 - x) + (1/2).log (1 + x) + C
I = (1/2) (log(1 + x) - log(1 - x)) + C
I = (1/2) log [ (1 + x) / (1 - x) ] + C

Answer 6

ok there is formulae for that use ∫ 1/(a²-x²) dx which ans. 1/2a log(a+x/a-x)

Answer 7

I have a question, how are you able to write all those mathematical symbols?

Answer 8

http://mathworld.wolfram.com/topics/DefiniteIntegrals.html