Find $\int {x\sqrt{4+x}}\,dx$.

Solution Substitution I  Let $u = 4 + x$. Then $du = dx.$ Since $u=4+x,$ $x=u-4.$ Substituting gives $$\begin{eqnarray*} \int {x\sqrt{4+x}}\,dx \underset{\underset{{\color{#0066A7}{\hbox{\(u=x+4\)}}}}{\color{#0066A7}{\uparrow }}} =&& \int \underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{( u-4) }} \underset{\underset{{\color{#0066A7}{\hbox{\(4+x\)}}}}{\color{#0066A7}{\uparrow }}}{{\sqrt{u}}}\,\,\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{du}}\, =\int {({u^{3/2}-4u^{1/2}})}\,du \\ &=& {\dfrac{{u^{5/2}}}{{{\dfrac{{5}}{{2}}}}}}-4\cdot {\dfrac{{u^{3/2}}}{{{\dfrac{{3}}{{2}}} }}}+C \\ &=&{\dfrac{{2({4+x})^{5/2}}}{{5}}}-{\dfrac{{8({4+x})^{3/2}}}{{3}}}+C \end{eqnarray*}$$

Substitution II  Let $u = \sqrt{4\,{+}\,x}$, so $u^{2} = 4 + x$ and $x = u^{2} - 4$. Then $dx=2u\,du$ and $$\begin{eqnarray*} \int {x\sqrt{4+x}}\,dx & = &\int {(}\underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{{u^{2}-4}}}{)(u)(}\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{{2u\,du}}}{)= 2 \int {(u^{4}-4u^{2})\,du}}=2\left[ {\dfrac{{u^{5}}}{{5}}}-{\dfrac{{4u^{3}}}{{3}}}\right] +C \\ &=&\dfrac{2}{5} \big(\sqrt{4+x}\,\big) ^{5} -\dfrac{8}{3} \big(\sqrt{4+x}\,\big) ^{3}+C= \dfrac{2(4+x)^{5/2}}{5} -\dfrac{8(4+x)^{3/2}}{3}+C   \end{eqnarray*}$$