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:

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

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