Answer
See below.
Work Step by Step
Differentiate the RHS and see if it equals$\quad x^{3}\cos^{-1}5x $
$\displaystyle \frac{d}{dx}[\frac{x^{4}}{4}\cos^{-1}5x]=\frac{1}{4}[4x^{3}\cos^{-1}5x+x^{4}\cdot\frac{-1}{\sqrt{1-(5x)^{2}}}\cdot 5]$
$=x^{3}\displaystyle \cos^{-1}5x-\frac{5}{4}\cdot\frac{x^{4}}{\sqrt{1-25x^{2}}}$
$\displaystyle \frac{d}{dx}\left[ \frac{5}{4}\int\frac{x^{4}dx}{\sqrt{1-25x^{2}}}\right]=\frac{5}{4}\cdot\frac{x^{4}}{\sqrt{1-25x^{2}}}$
$\displaystyle \frac{d}{dx}[RHS]=x^{3}\cos^{-1}5x-\frac{5}{4}\cdot\frac{x^{4}}{\sqrt{1-25x^{2}}}+\frac{5}{4}\cdot\frac{x^{4}}{\sqrt{1-25x^{2}}},$
$=x^{3}\cos^{-1}5x$
Thus, the integration formula stands.
