how do i integrate (1+x^2)/(1-x^2) from 0 to 0.5

Question

zarkon · Answer

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

turingtest · Answer

I wrote$$\int{1\over1-x^2}dx-\int{x^2\over1-x^2}dx$$the first one is a known hyperbolic tangent integral isn't it?
I didn't want to say because I wasn't sure

turingtest · Answer

+ in between*

turingtest · Answer

the second is a trig sub for sure though...

turingtest · Answer

or long division...

zarkon · Answer

$$\frac{1+x^2}{1-x^2}=\frac{1}{x+1}-\frac{1}{x-1}-1$$

turingtest · Answer

oh yeah, that thing they call partial fractions !!
I guess it eluded me at the moment :(

anonymous · Answer

another way i saw of doing it would be to write 2-(1-x^2)/ (1-x^2)

anonymous · Answer

[2-(1-x^2)]/ (1-x^2)

turingtest · Answer

still leads to partial fractions that way
... actually that is what I meant by long division

turingtest · Answer

...sort of

anonymous · Answer

yeah i guess. thanks for the help

amistre64 · Answer

1+2x^2+2x^4+2x^6 .....
          -------------
1-x^2 ) 1+x^2
            (1-x^2)
            --------
               2x^2
              (2x^2-2x^4)
              ------------
                         2x^4
                          ..........

$$\int   1+2x^2+2x^4+2x^6 ...\ dx$$

$$x+\frac{2}{3}x^3+\frac{2}{5}x^5+\frac{2}{7}x^7 ...$$

amistre64 · Answer

$$\sum_{n=0}^{inf}\frac{2(0.5)^{2n+1}}{2n+1}$$