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.
