Question

How do we determine the value of `int x/sqrt(x - 2) dx`?

Answer 1

The integral equals `2x(x - 2)^(1/2) - 4/3(x - 2)^(3/2) + C`

Explanation:

First rewrite the integral as a product.

`int x/sqrt(x - 2)dx = intx/(x - 2)^(1/2)dx = intx(x - 2)^(-1/2)dx`

Now let `u = x` and `dv = (x - 2)^(-1/2)dx`. We instantly know that `du = dx`. To integrate `dv`, we need to make a substitution. Let `n = x- 2`. Then `dn = dx`.

`int(x - 2)^(-1/2)dx = intn^(-1/2)dn = 2n^(1/2) = 2(x - 2)^(1/2)`

So, to summarize:

`{(u = x), (du = dx), (dv = (x - 2)^(-1/2)dx), (v = 2(x - 2)^(1/2)):}`

We can add the constant of integration at the end.

Now use the above values in the integration by parts formula, which is:

`intudv = uv - intvdu`

`intx(x - 2)^(-1/2)dx = x(2(x - 2)^(1/2)) - int2(x - 2)^(1/2)dx`

`intx(x - 2)^(-1/2)dx= 2x(x - 2)^(1/2) - 2int(x - 2)^(1/2)`

We solve this using another substitution. Let `m = x- 2`. Then `dm = dx`.

`intm^(1/2)dm = 2/3m^(3/2) = 2/3(x - 2)^(3/2)`

Therefore, we have:

`intx(x - 2)^(-1/2)dx = 2x(x - 2)^(1/2) - 4/3(x - 2)^(3/2) + C`

Hopefully this helps!

Answer 2

`=2/3 sqrt(x-2) (x + 4) + C`

Explanation:

`int (x)/(sqrt(x-2)) dx`

Using a substitution, we could say that: `u = sqrt(x-2)` so that `x = u^2 + 2 = implies dx = 2 u du`

The integration then becomes:

`int (u^2 + 2)/(u) * 2 u du`

`= 2 int u^2 + 2 du = (2 u^3)/3 + 4u + C`

If we undo the sub, we have:

`= (2 (sqrt(x-2))^3)/3 + 4 sqrt(x-2) + C`

`=sqrt(x-2) ( (2 x- 4)/3 + 4) + C`

`=sqrt(x-2) ( (2 x)/3 + 8/3) + C`

`=2/3 sqrt(x-2) (x + 4) + C`

With IBP we can recognise that:

`int (x)/(sqrt(x-2)) dx = int x * 1/(sqrt(x-2)) dx = int x * (2 sqrt(x-2))^prime dx`

[this is because `(2 sqrt(x-2))^prime = 1/(sqrt(x-2))`]

`= 2 xsqrt(x-2) - int (x)^prime 2 sqrt(x-2) dx`

`= 2 xsqrt(x-2) - 2 int sqrt(x-2) dx`

`= 2 xsqrt(x-2) - 2 * 2/3 (x-2)^(3/2) + C`

`= sqrt(x-2) ( 2 x - 2 * 2/3 (x-2) ) + C`

`= sqrt(x-2) ( (2x)/3 + 8/3) + C`

`=2/3 sqrt(x-2) (x + 4) + C`

Same thing!