@@chirayu_jain another way is to differentiate x*ln(x) then you Get 1+ln(x) so (d/dx)(x*ln(x))=1+ln(x) Now integrating both sides, X*ln(x)=integral ln(x) +x+c. Substraction x on both sides, now -c is just another constant, so we get, Integral ln(x) =x*ln(x)-x+c.
Other ways to integrate ln?
Take x = u + 1
Then use the Maclaurin series
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@@chirayu_jain another way is to differentiate x*ln(x) then you Get 1+ln(x) so (d/dx)(x*ln(x))=1+ln(x)
Now integrating both sides,
X*ln(x)=integral ln(x) +x+c.
Substraction x on both sides, now -c is just another constant, so we get,
Integral ln(x) =x*ln(x)-x+c.
@@chirayu_jain not Maclaurin, it needs to be centered at a interval so we can't always use it.