How do you evaluate the integral #int 1/((1+x^2)(2+x^2))dx#?

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Answer 1 · 2017-01-11T11:00:49

#int (dx)/((1+x^2)(2+x^2)) = arctan x -1/sqrt(2) arctan (x/sqrt(2))+C#

Simplify the integral using partial fractions:

#1/((1+x^2)(2+x^2)) = A/(1+x^2) +B/(2+x^2)#

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

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

#{ color(white)([)color(black)((A+B=0), (2A+B =1))]#

#{ color(white)([)color(black)((A= -B), (2A+B =1))]#

#{ color(white)([)color(black)((A= -B), (2A-A =1))]#

#{ color(white)([)color(black)((A= 1), (B=-1))]color(black)#

So:

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

Let's rewrite the second addendum as:

#int (dx)/(2+x^2) =1/sqrt(2) int (d(x/sqrt(2)))/(1+(x/sqrt(2))^2) #

and we have:

#int (dx)/((1+x^2)(2+x^2)) = arctan x -1/sqrt(2) arctan (x/sqrt(2))+C#