Question

Find the integral, using techniques from this or the previous chapter.
∫3x/√x-2 dx.

Answer 1

`2(x-2)^{(3)/(2)}+12√(x-2)+C`

Explanation:

Given:

The expression to integrate is given as:

`\int (3x)/(√(x-2))\ dx`

Now, in order to integrate it, we apply the method of substitution.

Let
`x-2=t^2`

`x=t^2+2`

Differentiating with respect to 't' on both sides, we get:

`(dx)/(dt)=2t+0\\(dx)/(dt)=2t\\dx=2tdt`

Replace
`dx` by
`2tdt`,
`x-2\ by\ t^2` and
`x` by
`t^2+2`. This gives,

`\int (3(t^2+2))/(√(t^2))\ 2tdt\\\\6\int ((t^2+2)t)/(t)\ dt\\\\6\int (t^2+2)\ dt\\\\6[(t^3)/(3)+2t]+C\\\\2t^3+12t+C`

Replacing 't' by
`√(x-2)`, we get:

`=2(x-2)√(x-2)+12√(x-2)+C`

`2(x-2)^{(3)/(2)}+12√(x-2)+C`

Therefore, the integral is:

`\int (3x)/(√(x-2))\ dx=2(x-2)^{(3)/(2)}+12√(x-2)+C`

Where 'C' is the constant of integration.