Question

Integrate: (x^3)ln(x) dx

Answer 1

Ans is
\[\frac{1}{4}x^4\ln(x) - \frac{1}{16}x^4+c\]
Let's try by parts
I would usually say let u be the part that derivatives would go to 0, but that would leave dv = ln(x), which we can't integrate directly. So this time let u = ln(x) and dv = x^3

Answer 2

ok

Answer 3

\[\int x^3\ln{x}dx\quad,\quad y=\ln{x}\,,\,dy=\frac{dx}{x}\,,\,x=e^y\]
\[\int x^3\ln{x}dx=\int x^4\ln{x}\frac{dx}{x}=\int e^{4y}ydy=\frac{ye^{4y}}{4}-\frac{1}{4}\int e^{4y}dy=\]
\[=\frac{ye^{4y}}{4}-\frac{1}{16}e^{4y}+C=\frac{e^{4y}}{16}(4y-1)+C=\frac{x^4}{16}(4\ln{x}-1)+C\]