Note that the derivative of the denominator is:
#d/dx (x^2+2)^2 = 4x(x^2+2) = 4x^3+8x#
so we can split the numerator as:
#x^3+x+1 = 1/4 (4x^3+4x+4) = 1/4( 4x^3+8x)- 1/4(4x-4)#
#x^3+x+1 = 1/4 (4x^3+4x+4) = 1/4( 4x^3+8x)- x +1#
Then using the linearity of the integral:
#int_0^1 (x^3+x+1)/(x^2+2)^2dx = 1/4 int_0^1 ( 4x^3+8x)/(x^2+2)^2dx -int_0^1 x /(x^2+2)^2 dx + int_0^1 dx /(x^2+2)^2 #
Now the first integral can be solved immediately:
#int_0^1 ( 4x^3+8x)/(x^2+2)^2dx = int_0^1 (d((x^2+2)^2))/(x^2+2)^2#
#int_0^1 ( 4x^3+8x)/(x^2+2)^2dx = [ln(x^2+2)^2]_0^1#
#int_0^1 ( 4x^3+8x)/(x^2+2)^2dx = 2(ln3-ln2)#
Similarly for the second integral:
#int_0^1 x /(x^2+2)^2 dx = 1/2 int_0^1(d(x^2+2))/(x^2+2)^2#
#int_0^1 x /(x^2+2)^2 dx = -1/2 [1/(x^2+2)]_0^1#
#int_0^1 x /(x^2+2)^2 dx =-1/2(1/3-1/2) = 1/12#
The third integral can be solved by substitution: let #x=sqrt2tant#, so that #dx = sqrt2sec^2t# and solve the indefinite integral:
#int dx/(x^2+2)^2 = sqrt2 int (sec^2tdt)/(2tan^2t+2)^2#
using the trigonometric identity:
#1+tan^2t = 1+ sin^2t/cos^2t =(cos^2t+sin^2t)/cos^2t = 1/cos^2t =sec^2t#
we have:
#int dx/(x^2+2)^2 = 1/(2sqrt2) int (sec^2tdt)/sec^4t#
#int dx/(x^2+2)^2 = 1/(2sqrt2) int cos^2t dt#
#int dx/(x^2+2)^2 = 1/(2sqrt2) int (1+cos 2t)/2 dt#
#int dx/(x^2+2)^2 = 1/(4sqrt2) int dt + 1/(8sqrt2) cos 2t d(2t)#
#int dx/(x^2+2)^2 = 1/(4sqrt2) t + 1/(8sqrt2) sin(2t) + C#
No to undo the substitution use the parametric formula:
#sin alpha = (2tan(alpha/2) )/(1+tan^2(alpha/2))#
So:
#int dx/(x^2+2)^2 = 1/(4sqrt2)arctan(x/sqrt2) + 1/(8sqrt2) (sqrt2x)/(1+x^2/2) + C#
#int dx/(x^2+2)^2 = 1/(4sqrt2)arctan(x/sqrt2) + 1/4 x/(x^2+2) + C#
and finally:
#int_0^1 dx/(x^2+2)^2 = [1/(4sqrt2)arctan(x/sqrt2) + 1/4 x/(x^2+2) ]_0^1#
#int_0^1 dx/(x^2+2)^2 = 1/(4sqrt2)arctan(1/sqrt2) + 1/12#
Putting it together:
#int_0^1 (x^3+x+1)/(x^2+2)^2dx = 1/2(ln3-ln2) -1/12 + 1/(4sqrt2)arctan(1/sqrt2) + 1/12#
and simplifying:
#int_0^1 (x^3+x+1)/(x^2+2)^2dx = 1/2(ln3-ln2) + 1/(4sqrt2)arctan(1/sqrt2)#
