Change of variable by substitution

Question

Given the function $$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$ You would need to have $u=\sqrt{x}-3$ and $du=\frac{1}{2 \sqrt{x}}$, when I use a online calculator it suggests to rewrite the numerator as $$\int \frac{2(u^2+6u+9)}{u}du $$ and I was wondering how do you arrive to this trinomial mathematically/analytical?

Answer 1

If $u=\sqrt{x}-3$, then $x=(u+3)^2$ and $dx=2(u+3)du$. Therefore $$\int \frac{\sqrt x}{\sqrt{x}-3}dx=\int \frac{u+3}{u}\cdot 2(u+3)du=\int \frac{2(u+3)^2}{u}du$$ My recommendation is always to write $x$ according to $u$, after perform the substitution.

Answer 2

If $u=\sqrt{x}-3,$ then $\sqrt{x}=u+3,$ and $dx=2\sqrt{x} du=2(u+3)du.$ Making these substitutions, $$\int\frac{\sqrt{x}}{\sqrt{x}-3}\, dx=\int\frac{u+3}{u}2(u+3)\, du=\int\frac{2(u^2+6u+9)}{u}\, du $$

Answer 3

Simple: 1) $\;\sqrt x=u+3$, and 2) $\;\mathrm d u=\dfrac{\mathrm dx}{2\sqrt x}$, so $$\mathrm dx=2\sqrt x\,\mathrm d u=2(u+3)\,\mathrm d u, \quad\text{so }\quad\sqrt x\,\mathrm dx=2(u+3)^2\mathrm du.$$