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

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!

#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!