How do you integrate #int x/sqrt(3 + x^2)dx# using trigonometric substitution?

Question

5218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T10:50:26

#int x/sqrt(3+x^2)*dx=sqrt(x^2+3)+C#

#int x/sqrt(3+x^2)*dx#

After using #x=sqrt3tany# and #dx=sqrt3*(secy)^2*dy# transforms, this integral became

#int sqrt3tany*(sqrt3*(secy)^2*dy)/(sqrt3*secy)#

=#int sqrt3secy*tany*dy#

=#sqrt3secy+C#

After using #tany=x/sqrt3# and #secy=sqrt(x^2+3)/sqrt3# inverse transforms, I found

#int x/sqrt(3+x^2)*dx=sqrt3*sqrt(x^2+3)/sqrt3+C=sqrt(x^2+3)+C#