Explanation
Consider the given integral.
$$I=\int{{{x}^{3}}\log xdx}$$
$$ I=\log x\left( \dfrac{{{x}^{4}}}{4} \right)-\int{\dfrac{1}{x}}\left( \dfrac{{{x}^{4}}}{4} \right)dx $$
$$ I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{1}{4}\int{{{x}^{3}}}dx $$
$$ I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{1}{4}\left( \dfrac{{{x}^{4}}}{4} \right)+C $$
$$ I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{{{x}^{4}}}{16}+C $$
Hence, this is the answer.
$$\int log_{10}xdx\\=\int(\dfrac{log_ex}{log_e10})dx\\=(\dfrac{1}{log_e10})\int logx\cdot1\cdot dx\\(\dfrac{1}{log_e10})[(logx)\cdot x-\int(\dfrac{1}{x}\cdot x dx)]\\=(\dfrac{d1}{log_e10})(xlogx-x)+c\\=x((\dfrac{log_ex}{log_e10})-(\dfrac{1}{log_e10}))+c\\=x(log_{10}x-log_{10}e)+C$$
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