How do you find the indefinite integral of #int x^2/(3-x^2)#?

1 Answer
Jan 23, 2017

#= - x + sqrt 3 tanh^(-1) (x/ (sqrt 3)) + C#

Explanation:

#int x^2/(3-x^2) color(red)(dx)#

#= int ((- 3 + x^2) + 3)/(3-x^2) dx#

#= int -1 + (3)/(3-x^2) dx#

#= -x + color(blue)( int (3)/(3-x^2) dx) star#

For the blue bit, we will use the hyperbolic identity:

#1 - tanh^2 y = sech^2 y#

So we let #x = sqrt 3 tanh y implies dx =sqrt 3 sech^2 y dy# so the blue part of #star# becomes:

# int (3)/(3-3 tanh^2 y) sqrt 3 sech^2 y \ dy#

#= int (3)/(3-3 tanh^2 y) sqrt 3 sech^2 y \ dy#

#= int 1/sech^2 y sqrt 3 sech^2 y \ dy#

#= sqrt 3 int dy#

#= sqrt 3 tanh^(-1) (x/ (sqrt 3)) + C#

Feeding this into #star#

#implies - x + sqrt 3 tanh^(-1) (x/ (sqrt 3)) + C#