How do you integrate #int x/sqrt(x^2+9)dx#?

2 Answers
Sep 17, 2015

substitute x = #3 tantheta#
#(x^2+9)#=#9tan^2theta+9#= #9sec^2theta#
#(x^2+9)^(1/2)# = #3sec theta#
#dx = 3 sec^2theta d theta#
substiuting
#int##(3tantheta)/(3sec theta)# #3 sec^2theta d theta#
=#3int# #sectheta tanthetad theta#
=#3sectheta+c#
#sectheta = sqrt ##(1+tan^2theta)#=#(root##1+(x/3)^2)#
#sqrt(9+x^2)/3#
=>#3sectheta#=#sqrt(9+x^2#
the final answer is #sqrt(9+x^2)+c#
it can be done in other way
#x is derrivative of x^2/2#
it can be wriiten as #d(x^2)/2#
so #int x/sqrt(x^2+9)dx# bexmes #1/2int1/sqrt(x^2+9)d(x^2)#
=#1/2(2sqrt(x^2+9))#+#c#
=#sqrt(x^2+9)+c#

Sep 18, 2015

#sqrt(x^2+9)+C#

Explanation:

substitution:

#x^2+9=t => 2xdx=dt => xdx=dt/2#

#intx/sqrt(x^2+9)dx=intdt/2 1/sqrtt=1/2intt^(-1/2)dt=#

#=1/2 t^(1/2)/(1/2)+C=sqrtt+C=sqrt(x^2+9)+C#