Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 33

int1/(4+4x^2+x^4)dx
Let's rewrite the integrand as :
=int1/((x^2)^2+2(2)(x^2)+2^2)dx
=int1/(x^2+2)^2dx
Apply integral substitution:x=sqrt(2)tan(u),u=arctan(x/sqrt(2))
dx=sqrt(2)sec^2(u)du
=int(sqrt(2)sec^2(u))/((sqrt(2)tan(u))^2+2)^2du
=int(sqrt(2)sec^2(u))/(2tan^2(u)+2)^2du
=int(sqrt(2)sec^2(u))/(2^2(tan^2(u)+1)^2)du
Take the constant out,
=sqrt(2)/2^2int(sec^2(u))/(tan^2(u)+1)^2du
Use the identity:1+tan^2(x)=sec^2(x)
=sqrt(2)/4int(sec^2(u))/(sec^2(u))^2du
=sqrt(2)/4int1/(sec^2(u))du
=sqrt(2)/4intcos^2(u)du
Use the trigonometric identity:cos^2(x)=(1+cos(2x))/2
=sqrt(2)/4int1/2(1+cos(2u))du
Take the constant out,
=sqrt(2)/8int(1+cos(2u))du
Apply the sum rule,
=sqrt(2)/8{int1du+intcos(2u)du}
Apply the common integral:intcos(x)dx=sin(x)
=sqrt(2)/8{u+1/2sin(2u)}
Substitute back u=arctan(x/sqrt(2))
=sqrt(2)/8{arctan(x/sqrt(2))+1/2sin(2arctan(x/sqrt(2)))}
=sqrt(2)/8{arctan(x/sqrt(2))+1/2(2sin(arctan(x/sqrt(2)))cos(arctan(x/sqrt(2))))}
=sqrt(2)/8{arctan(x/sqrt(2))+sin(arctan(x/sqrt(2)))cos(arctan(x/sqrt(2)))}
Use the identities:sin(arctan(x))=x/sqrt(1+x^2),cos(arctan(x))=1/sqrt(1+x^2)
=sqrt(2)/8{arctan(x/sqrt(2))+x/sqrt(x^2+2)sqrt(2)/(sqrt(x^2+2))}
=sqrt(2)/8{arctan(x/sqrt(2))+(sqrt(2)x)/(x^2+2)}
=1/8{sqrt(2)arctan(x/sqrt(2))+2x/(x^2+2)}
Add a constant C to the solution,
=1/8(sqrt(2)arctan(x/sqrt(2))+(2x)/(x^2+2))+C