We can write the question as `int(dx)/((3^2+x^2)^(3//2))`

It's now in the form of the second substitution suggestion given above, that is:

For `sqrt(a^2+x^2)`, use `x=a\ tan\ theta`,

with `a=3`.

So we'll put `x=3\ tan\ theta` and this gives `dx=3\ sec^2 theta\ d theta`

We make the first substitution and simplify the denominator of the question before proceeding to integrate.

We'll need to use the following:

`(a^2)^(3//2) = a^3`.

Here's a number example demonstrating this expression:

`9^(3//2) = (sqrt9)^3 = 3^3 = 27`

This is a well-known trigonometric identity:

`tan^2 θ + 1 = sec^2 θ`

So we have:

`(x^2+9)^(3//2)=((3\ tan\ theta)^2+9)^(3//2)`

`=(9\ tan^2 theta+9)^(3//2)`

`=(9[tan^2 theta+1])^(3//2)`

`=9^(3//2)[tan^2 theta+1]^(3//2)`

`=27[sec^2 theta]^(3//2)`

`=27[sec\ theta]^3`

`=27\ sec^3 theta`

Now, substituting

`dx=3\ sec^2 theta\ d theta`

and

`(x^2+9)^(3//2)=27\ sec^3 theta`

into the given integral gives us:

`int(dx)/((x^2+9)^(3//2))=int(3\ sec^2 theta\ d theta)/(27\ sec^3 theta)`

`=1/9int(d theta)/(sec\ theta)`

`=1/9int cos\ theta\ d theta`

`=1/9 sin\ theta+K`

We now need to get our answer in terms of x (since the question was in terms of x).

Since we let `x = 3\ tan θ`, we get

`tan\ theta=x/3`

and we can draw a triangle to find the value of sin θ :

Hence, we notice that

`sin\ theta=x/sqrt(x^2+9)`

Therefore, we can conclude that the answer for our integral is `1/9` times this last expression.:

`int(dx)/((x^2+9)^(3//2))=1/9sin\ theta+K`

`=1/9(x/(sqrt(x^2+9)))+K`

`=x/(9sqrt(x^2+9))+K`

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