Question

Integral of 1/(1-y2) Integral of 1/(1-y2) @Mathematics

Answer 1

∫dy /(y^2 + 2y)

take out y as common factor from denominator.

=>∫dy /y(y+2)

resolving 1/y(y+2) into partial fractions

let 1/y(y+2) = A/y + B/(y+2)

1 = A(y+2) + By

1 = y(A+B) + 2A

comparing coefficients from LHS and RHS

A+B = 0 and

2A = 1

A = 1/2 and B = -1/2

∫dy/y(y+2) = 1/2∫dy/y - 1/2∫dy/(y+2)

=>(1/2)ln(y) - (1/2)ln(y+2) + c

=>(1/2)[lny - ln(y+2)] +c

=>(1/2)[lny/(y+2)] + c (since lna - ln b = ln(a/b))

Answer 2

1/2ln(1+y)+1/2ln(1-y)+c

Answer 3

http://www.wolframalpha.com/input/?i=Integral+of+1%2F%281-y2%29
In order to do this problem you must recognize that
1/(1-y2)=htan^-1(y) which I frankly did not know myself.

Answer 4

INT 1/(1-y2)dy
write as partial fractions
and integrate each fraction

Answer 5

Oh yeah, that should work too!

Answer 6

i have to go - i know thtas the way to do it but i got in a mess!!