Practicing Success
If $I_{n}=\int(\ln x)^n d x$ then $I_n+n I_{n-1}$ is equal to |
$x(\ln x)^{n-1}$ $x(\ln x)^n$ $n x(\ln x)^n$ none of these |
$x(\ln x)^n$ |
$I_{n}=\int(\ln x)^n d x$ $I_n=x(\ln x)^n-n \int(\ln x)^{n-1} d x$ $I_n=x(\ln x)^n-n I_{n-1}$ $I_n+n I_{n-1}=x(\ln x)^n$ Hence (2) is the correct answer. |