Question

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

Answer 1

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`

Answer 2

`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`