Lately, I have done the integration by substitution that involves the square roots, and I do the manipulation algebraically when there's a square root like this.
So, I have $\displaystyle\int x^3\sqrt{16x^2-4}\,\mathrm dx$. Then I do the substitution normally like this:
Let \begin{align*} u&={16x^2-4} \\\color{Green} {\dfrac{\mathrm{d}u}{\mathrm{d}x}}&=\color{Green}{32x} \implies \color{Green}{\mathrm du}=\color{Green}{32x}\,\color{Green}{\mathrm {d}x}\implies \color{Green}{\mathrm dx} =\color{Green}{\dfrac{1}{32x}}\,\color{Green}{\mathrm du} \end{align*} Then my manipulation algebraically is by doing this: \begin{align*} u&=16x^2-4\\ \color{Green}{4-}u&= 16x^2\color{Green} {-4+4}\\ 4-u &= 16x^2 \tag*{This part make me confused should I taking $\sqrt{\alpha^m}$ or not }\\4-u&=\frac{1}{16x^2} \end{align*} Here for the integral. \begin{align*}\displaystyle &\implies\int x^3\sqrt{\color{Green} {16x^2-4}} \,\mathrm dx \\ &\implies\int \color{Green}{x^3}\sqrt u\cdot\left(\color{Green} {\frac{1}{32x}} \right)\,\mathrm du\\ &\implies \int\left( \color{Green}{x^2} \right)(\color{Red} x)\sqrt{u}\cdot\left( \frac{1}{32\color{Red} x} \right)\left( \color{Green}{\frac{1}{16}} \right)\mathrm du \\ &\implies\frac{1}{512}\int\left( 4-u \right)\sqrt u\,\mathrm du\\ &\implies\frac{1}{512}\int4\sqrt u-u\sqrt u\,\mathrm du\\ &\implies\frac{1}{512}\int4u^{\frac12}-u^{\frac32} \,\mathrm du\\ &\implies\frac{1}{\color{Red}{512}}\left[\color{Red} 4\left( \frac{2u^{\frac32}}{3} \right)-\frac{2u^{\frac52}}{5} \right] \\ &\implies\frac{1}{128}\left[ \frac{2\left( 16x^2-4 \right)^{\frac32}}{3}-\frac{2\left( 16x^2-4 \right)^{\frac52}}{5} \right] \\ &\implies\frac{\left(16x^2-4\right)^{\frac32}}{192}+\frac{\left( 16x^2-4 \right)^{\frac{5}{2}}}{320}+C \end{align*}
Did I make a mistake on those steps I made?
$$ \int x^3 \sqrt{16x^2-4} dx = \int x^3 \sqrt u \frac{du}{32x} $$ $$ = \frac{1}{32}\int x^2 \sqrt u du = \frac{1}{32}\int \frac{u+4}{16} \sqrt u du $$ $$ = \frac{1}{512}\int [u^{3/2}+4u^{1/2}] du = \frac{1}{512}[\frac{2}{5} u^{5/2}+\frac{8}{3} u^{3/2}] $$ $$ = \frac{1}{1280}(16x^2-4)^{5/2}+\frac{1}{192}(16x^2-4)^{3/2} +C $$