The answer is: #xsqrt(x^2+3)+3arc sinh(x/sqrt3)+c#.
The integral can be written:
#intsqrt(4(3+x^2))dx=2intsqrt(3+x^2)dx=(1)#
Since:
#x=sqrt3sinhtrArrdx=sqrt3coshtdt#, than:
#(1)=2intsqrt(3+3sinh^2t)*sqrt3coshtdt=#
#=2intsqrt3sqrt(1+sinh^2t)sqrt3coshtdt=6intcosht*coshtdt=#
#=6intcosh^2tdt=(2)#.
Now remembering that:
#cosh(alpha/2)=sqrt((coshalpha+1)/2)#,
#(2)=6int(cosh2t+1)/2dt=6/2(1/2int2cosh2t+intdt)=#
#=3(1/2sinh2t+t)+c=(3)#
And now, remembering that:
#sinh2alpha=2sinhalphacoshalpha# and
#coshalpha=sqrt(sinh^2alpha+1)#,
than:
#(3)=3(1/2 2sinhtcosht+t)=#
#=3(x/sqrt3sqrt(x^2/3+1)+arc sinh(x/sqrt3))+c=#
#=3(x/sqrt3sqrt(x^2+3)/sqrt3+arc sinh(x/sqrt3))+c=#
#=xsqrt(x^2+3)+3arc sinh(x/sqrt3)+c#.
