Find the integral of 4/(1-x^2) dx:

The first thing to notice here is that the denominator of the integrand is a case of 'difference of two squares'. The integral, which I will call I, can be rewritten as the integral of 4/((1+x)(1-x)) dx. If you expand the brackets you will find the denominator gives (1-x²) as in the question. Now we can apply partial fractions to further simplify I. 4/((1+x)(1-x)) = A/(1+x) + B/(1-x) Multiply both sides by (1+x)(1-x). 4 = A(1-x) + B(1+x) Sub in x = -1 to eliminate B. 4 = 2A so A=2Sub in x = 1 to eliminate A. 4 = 2B so B=2 Now we can integrate using the fact that the integral of 1/y dy = lny + c. The integral of (2/(1+x) +2/(1-x))dx = 2ln(1+x) + 2ln(1-x) (-1) + c . There is a factor of -1 in the second term because it was (1-x). Factorise the 2 and use the subtraction of logs rule (lna - lnb = ln(a/b)), to give: I = 2ln((1+x)/(1-x)) + c