Question

Find the integrals:
∫30x^2/√(x-4) dx
u=x-4 and u=√(x-4)

Answer 1

I assume you're asked to compute

`\displaystyle \int (30x^2)/(√(x-4)) \, dx`

using both of the substitutions provided.

With
`u=x-4`, we have
`x=u+4` and
`dx=du`. Then

`\displaystyle \int (30x^2)/(√(x-4)) \, dx = \int (30(u+4)^2)/(\sqrt u) \, du \\\\ ~~~~~~~~ = 30 \int (u^2 + 8u + 16)/(\sqrt u) \, du \\\\ ~~~~~~~~ = 30 \int \left(u^(3/2) + 8u^(1/2) + 16u^(-1/2)\right) \, du \\\\ ~~~~~~~~ = 30 \left(\frac25 u^(5/2) + \frac{16}3 u^(3/2) + 32 u^(1/2)\right) + C \\\\ ~~~~~~~~ = 12 u^(5/2) + 160 u^(3/2) + 960 u^(1/2) + C \\\\ ~~~~~~~~ = 12 (x-4)^(5/2) + 160 (x-4)^(3/2) + 960 (x-4)^(1/2) + C \\\\ ~~~~~~~~ = 4 √(x-4) \left(3 (x-4)^2 + 40 (x-4) + 240\right) + C \\\\ ~~~~~~~~ = \boxed{4 √(x-4) \left(3x^2 + 16x + 128\right) + C}`

With
`u=√(x-4)`, we have

`u^2 = x-4 \implies x^2 = (u^2+4)^2`

and
`2u\,du=dx`. Then

`\displaystyle \int (30x^2)/(√(x-4)) \, dx = \int \frac{60u \left(u^2+4\right)^2}u \, du \\\\ ~~~~~~~~ = 60 \int \left(u^4 + 8u^2 + 16\right) \, du  \\\\ ~~~~~~~~ = 60 \left(\frac15 u^5 + \frac83 u^3 + 16u\right) + C  \\\\ ~~~~~~~~ = 12 (x-4)^(5/2) + 160 (x-4)^(3/2) + 960 (x-4)^(1/2) + C \\\\ ~~~~~~~~ = 4 √(x-4) \left(3 (x-4)^2 + 40 (x-4) + 240\right) + C \\\\ ~~~~~~~~ = \boxed{4√(x-4) \left(3x^2 + 16x + 128\right) + C}`