$\int x^x\left(1+\log _e x\right) d x$ is equal to |
$x^x \log _e x+C$ $e^{x^x}+C$ $x^x+C$ none of these |
$x^x+C$ |
The correct answer is Option (3) → $x^x+C$ $I=\int x^x\left(1+\log x\right) d x$ and, $x^x=e^{x\log x}$ $⇒I=\int e^{x\log x}(1+\log x)dx$ ...(1) let, $t=x\log x$ $⇒\frac{dt}{dx}=\log x+1$ Substituting this in (1), $I=\int e^tdt$ $=e^t=e^{x\log x}+C$ $=x^x+C$ |