Question #a0e4b

1 Answer
May 19, 2017

#(1/4)ln(abs(x-2))-(1/4)(ln(abs(x+2)))+C#

Explanation:

I will assume you meant #int (dx/(x^2-4))#:

Rewrite #x^2-4# as #(x-2)(x+2)#:

#int (dx/((x-2)(x+2)))#

Rewrite the fraction as #A/(x-2) + B/(x+2)#. Solve for #A# and #B# by multiplying the three fractions by #(x-2)(x+2)#:

#1=A(x+2)+B(x-2)#

Plug in #2# for #x# to solve for #A#:

#1=A(4)#

#A=1/4#

Plug in #-2# for #x# to solve for #B#:

#1=-4B#

#B=-1/4#

Now we have:

#int (1/4)(1/(x-2)) + (-1/4)((1/(x+2)))#

This, we can solve by integrating:

#(1/4)ln(abs(x-2))-(1/4)(ln(abs(x+2)))+C#

If you want, you can write this as one log with log rules:

#(1/4)ln(abs((x-2)/(x+2)))+C#