Practicing Success
$\int x^x(1+\log x) dx$ is equal to : |
$x^x-c$ $x^x+c$ $x \log x+c$ None of these |
$x^x+c$ |
Let $I=\int x^x(1+\ln x) dx$ Let $x^{x}=t \Rightarrow x \ln x=\ln t$ $\Rightarrow\left(x . \frac{1}{x}+\ln x . 1\right)=\frac{1}{t} \frac{d t}{d x}$ $\Rightarrow dx(1+\ln x) x^{x}=dt$ ∴ $I=\int d t \Rightarrow I=t+c$ $I=x^{x}+c$ Hence (2) is the correct answer. |