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)=(root1+(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