Answer
$$\int^2_1 x\ln xdx=2\ln2-\frac{3}{4}$$
Work Step by Step
$$A=\int^2_1 x\ln xdx$$
Take $u=\ln x$ and $dv=x dx$
We then have $du=\frac{1}{x}dx$ and $v=\frac{x^2}{2}$
Apply the formula $\int^b_a udv= uv]^b_a-\int^b_a vdu$, we have $$A=\Big(\ln x\times\frac{x^2}{2}\Big)\Bigg]^2_1-\int^2_1\frac{x^2}{2}\times\frac{1}{x}dx$$ $$A=2\ln2-\frac{1}{2}\ln1-\int^2_1\frac{x}{2}dx$$ $$A=2\ln2-\Big(\frac{x^2}{4}\Big)\Bigg]^2_1$$ (since $\ln1=0$) $$A=2\ln2-\Big(1-\frac{1}{4}\Big)$$ $$A=2\ln2-\frac{3}{4}$$
