Question

How do you integrate ? 1/(x^2+9)^(1/2)

Answer 1

`y=int1/sqrt(x^2+9) dx`

put `x=3 tant` `rArr t=tan^-1(x/3)`
Hence, `dx= 3sec^2tdt`

`y=int(3sec^2t)/sqrt(9tan^2t+9) dt`

`y=int(sec^2t)/sqrt(tan^2t+1) dt`

`y=int(sec^2t)/sqrt(sec^2t) dt`

`y=int(sec^2t)/(sect) dt`

`y=int(sect) dt`

`y=ln|sec t + tan t| + C`

`y=ln|sec( tan^-1(x/3) )+ tan (tan^-1(x/3))| + C`

`y=ln|sec( tan^-1(x/3) )+ x/3)| + C`

`y=ln|sqrt(1+x^2/9 )+ x/3| + C`

Answer 2

We know that,
`int1/sqrt(X^2+A^2) dX=ln|X+sqrt(X^2+A^2)|+c`
So,
`I=int1/(x^2+9)^(1/2) dx=int1/sqrt(x^2+3^2)dx`
`=>I=ln|x+sqrt(x^2+9)|+c`

Explanation:

`II^(nd)` method : Trig. subst.

`I=int1/(x^2+9)^(1/2) dx`

Take, `x=3tanu=>dx=3sec^2udu`

`and color(blue)(tanu=x/3`

So,

`I=int1/(9tan^2u+9)^(1/2) 3sec^2udu`

`=int(3sec^2u)/((9sec^2u)^(1/2))du`

`=int(3sec^2u)/(3secu)du`

`=intsecudu`

`=ln|secu+tanu|+c`

`=ln|sqrt(tan^2u+1)+tanu|+c` , where, `color(blue)(tanu=x/3`

`:.I=ln|sqrt(x^2/9+1)+x/3|+c`

`=ln|sqrt(x^2+9)/3+x/3|+c`

`=ln|(sqrt(x^2+9)+x)/3|+c`

`=ln|sqrt(x^2+9)+x|-ln3+c`

`=ln|x+sqrt(x^2+9)|+C,where, C=c-ln3`