CUET Preparation Today
The value of $\int x \log x(\log x-1) d x$ is equal to |
$2(x \log x-x)^2+C$ $\frac{1}{2}(x \log x-x)^2+C$ $(x \log x)^2+C$ $\frac{1}{2}(x \log x)^3+C$ |
$\frac{1}{2}(x \log x-x)^2+C$ |
Let $I=\int x \log x(\log x-1) d x=\int \log x(x \log x-x) d x$ $\Rightarrow I=\int(x \log x-x) d(x \log x-x)=\frac{(x \log x-x)^2}{2}+C$ |
