$int_{-1}^{e}xln xdx$

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = \ln x$ and $\mathrm{d}v=x\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v=\frac{1}{2}x^{2}$.

Step2: Apply the integration - by - parts formula

$\int_{-1}^{e}x\ln x\mathrm{d}x=\left[\frac{1}{2}x^{2}\ln x\right]{-1}^{e}-\int{-1}^{e}\frac{1}{2}x^{2}\cdot\frac{1}{x}\mathrm{d}x$. But the function $y = \ln x$ is not defined for $x=-1$ in the real - number system. The domain of the natural logarithm function $y = \ln x$ is $(0,+\infty)$. So, we consider the integral $\int_{1}^{e}x\ln x\mathrm{d}x$. Applying integration - by - parts again for $\int_{1}^{e}x\ln x\mathrm{d}x=\left[\frac{1}{2}x^{2}\ln x\right]{1}^{e}-\int{1}^{e}\frac{1}{2}x\mathrm{d}x$. First, evaluate $\left[\frac{1}{2}x^{2}\ln x\right]{1}^{e}$: When $x = e$, $\frac{1}{2}x^{2}\ln x=\frac{1}{2}e^{2}\ln e=\frac{1}{2}e^{2}$. When $x = 1$, $\frac{1}{2}x^{2}\ln x=\frac{1}{2}\times1^{2}\ln1 = 0$. Second, evaluate $\int{1}^{e}\frac{1}{2}x\mathrm{d}x=\frac{1}{2}\times\frac{1}{2}x^{2}\big|_{1}^{e}=\frac{1}{4}(e^{2}-1)$.

Step3: Calculate the result

$\int_{1}^{e}x\ln x\mathrm{d}x=\frac{1}{2}e^{2}-\frac{1}{4}(e^{2}-1)=\frac{1}{2}e^{2}-\frac{1}{4}e^{2}+\frac{1}{4}=\frac{1}{4}e^{2}+\frac{1}{4}$.

Answer:

$\frac{1}{4}e^{2}+\frac{1}{4}$