Question

How do you use substitution to integrate `x/sqrt(x+4) dx`?

Answer 1

The process is outlined below.

Explanation:

We are to evaluate, `int x/(x + 4)^(1/2)*dx`.

We shall accomplish this by substitution.

Let, `(x + 4) = t`
`implies dx = dt` (By simple differentiation).

Thus, the integral becomes,

`int x/(x +4)^(1/2)*dx = int (t - 4)/t^(1/2)*dt`

`= int t^(1/2)*dt - int 4t^(-1/2)*dt`

`=2/3 t^(3/2) - 8t^(1/2) + C`, where `C` is the integration constant.

In terms of `x`, the integral may be now written as,

`int x/(x + 4)^(1/2)*dx =2/3 (x + 4)^(3/2) - 8(x + 4)^(1/2) + C`.

Answer 2

`int x/(sqrt(x+4)) "d"x = 2/3 (x+4)sqrt(x+4) - 8 sqrt(x+4) + C`

Explanation:

`int x/(sqrt(x+4)) "d"x = int (x+4-4)/(sqrt(x+4)) "d"x`,
`int x/(sqrt(x+4)) "d"x = int (x+4)/((x+4)^(1/2))-4/((x+4)^(1/2)) "d"x`,
`int x/(sqrt(x+4)) "d"x = int (x+4)^(1/2)-4(x+4)^(-1/2) "d"x`,
`int x/(sqrt(x+4)) "d"x = 2/3(x+4)^(3/2)-8(x+4)^(1/2) + C`