The value of $\int x \log x(\log x-1) d x$ is equal to
Answer & explanation
Correct answer: option 2
Let
$I=\int x \log x(\log x-1) d x=\int \log x(x \log x-x) d x$
$\Rightarrow I=\int(x \log x-x) d(x \log x-x)=\frac{(x \log x-x)^2}{2}+C$
The value of $\int x \log x(\log x-1) d x$ is equal to
Correct answer: option 2
Let
$I=\int x \log x(\log x-1) d x=\int \log x(x \log x-x) d x$
$\Rightarrow I=\int(x \log x-x) d(x \log x-x)=\frac{(x \log x-x)^2}{2}+C$