Question

∫ x/(x-3)^.5 dx

Answer 1

2/3(x-3)^0.5(x+6)+constant

Answer 2

The indefinite integral

Answer 3

can you explain

Answer 4

let u=x-3
du=dx

\[\huge{\int\limits{\frac{u+3}{\sqrt{u}}du}}=\int\limits{\frac{u}{\sqrt{u}}du}+\int\limits{\frac{3}{\sqrt{u}}du}\]

Answer 5

Let (x-3) = t, hence, x = t + 3

so, dx = dt

Thus the integral transforms to:
\[\int\frac{t+3}{\sqrt t}dt\]

Answer 6

\[=\int{\sqrt t dt + \int 3 t^{\frac{-1}2}}dt\]


Can you do this now?

Answer 7

yes i think i got it thanx