Integral dx/x+√x^2+x+1.?

Let

#I=intdx/(x+sqrt(x^2+x+1))#

Rationalize:

#I=intdx/(x+sqrt(x^2+x+1))*(x-sqrt(x^2+x+1))/(x-sqrt(x^2+x+1))#

Simplify:

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

Rearrange:

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

Complete the square in the square root:

#I=1/2intsqrt((2x+1)^2+3)/(x+1)dx-(x-ln|x+1|)#

Apply the substitution #2x+1=sqrt3tantheta#:

#I=1/2int(sqrt3sectheta)/(sqrt3/2tantheta+1/2)(sqrt3/2sec^2thetad theta)-x+ln|x+1|#

Simplify:

#I=3/2intsec^2theta/(sqrt3tantheta+1)secthetad theta-x+ln|x+1|#

Since #sec^2theta=tan^2theta+1#:

#I=1/2int(sqrt3tantheta-1+4/(sqrt3tantheta+1))secthetad theta-x+ln|x+1|#

Rearrange:

#I=1/2int(sqrt3secthetatantheta-sectheta)d theta+2int1/(sqrt3sintheta+costheta)d theta-x+ln|x+1|#

Apply the appropriate double-angle trigonometric identities:

#I=1/2(sqrt3sectheta-ln|sqrt3sectheta+sqrt3tantheta|)+2intsec^2(theta/2)/(2sqrt3tan(theta/2)+1-tan^2(theta/2))d theta-x+ln|x+1|#

Complete the square in the denominator:

#I=1/2(sqrt3sectheta)-1/2ln|sqrt3sectheta+sqrt3tantheta|+2intsec^2(theta/2)/(4-(tan(theta/2)-sqrt3)^2)d theta-x+ln|x+1|#

Apply partial fraction decomposition:

#I=1/2(sqrt3sectheta)-1/2ln|sqrt3sectheta+sqrt3tantheta|+1/2int(1/(2+tan(theta/2)-sqrt3)+1/(2-tan(theta/2)-sqrt3))sec^2(theta/2)d theta-x+ln|x+1|#

Integrate directly:

#I=1/2(sqrt3sectheta)-1/2ln|sqrt3sectheta+sqrt3tantheta|+(ln|2+tan(theta/2)-sqrt3|-ln|2-tan(theta/2)-sqrt3|)-x+ln|x+1|#

Apply the trigonometric identity #tan(theta/2)=tantheta/(1+sectheta)#:

#I=1/2(sqrt3sectheta)-1/2ln|sqrt3sectheta+sqrt3tantheta|+ln|((2-sqrt3)(sqrt3+sqrt3sectheta)+sqrt3tantheta)/((2-sqrt3)(sqrt3+sqrt3sectheta)-sqrt3tantheta)|-x+ln|x+1|#

Reverse the substitution:

#I=1/2sqrt((2x+1)^2+3)-1/2ln|2x+1+sqrt((2x+1)^2+3)|+ln|((2-sqrt3)(sqrt3+sqrt((2x+1)^2+3))+(2x+1))/((2-sqrt3)(sqrt3+sqrt((2x+1)^2+3))-(2x+1))|-x+ln|x+1|#

Collect terms:

#I=(sqrt(x^2+x+1)-x)+ln|x+1|-1/2ln|2x+1+2sqrt(x^2+x+1)|+ln|((2-sqrt3)(sqrt3+2sqrt(x^2+x+1))+(2x+1))/((2-sqrt3)(sqrt3+2sqrt(x^2+x+1))-(2x+1))|#