How did we integrated RHS and git the answer in log?

$y (x^{2} - 1) = \int \frac{1}{x^{2} - 1} d x + C y (x^{2} - 1) = \frac{1}{2} \log |\frac{x - 1}{x + 1}| + C .$

Question

How did we integrated RHS and git the answer in log?

y   x 2   - 1   =   ∫ 1 x 2   - 1
  d x   +   C y   x 2   - 1   =
  1 2   log   x   - 1 x   +   1
  +   C

Varun Rawat · Accepted Answer

Let I = ∫dxx2 - 1=∫dxx + 1x - 1Let 1x + 1x - 1 = Ax+1 + Bx-1⇒1x + 1x - 1 = Ax - A + Bx + Bx + 1x - 1⇒1 = A+Bx + B-ACompring the coefficient of x on both sides, we getA + B = 0   ⇒ B = -AComparing the constants on both sides, we getB - A = 1⇒-A - A = 1⇒A = -12Now, B = 12Now, 1x + 1x - 1 = -12×1x+1 + 12×1x-1So, I = -12∫dxx+1 + 12∫dxx-1=12 logx - 1 - 12 logx + 1 + C=12 logx - 1x + 1 + C

How did we integrated RHS and git the answer in log? y x 2 - 1 = ∫ 1 x 2 - 1 d x + C y x 2 - 1 = 1 2 log x - 1 x + 1 + C .

How did we integrated RHS and git the answer in log?

$y (x^{2} - 1) = \int \frac{1}{x^{2} - 1} d x + C y (x^{2} - 1) = \frac{1}{2} \log |\frac{x - 1}{x + 1}| + C .$