Answer
$$\int^{4}_2(1+\ln t)t\ln tdt=30(\ln2)^2$$
Work Step by Step
$$A=\int^{4}_2(1+\ln t)t\ln tdt$$
Set $u=t\ln t$, which means $$du=(t'\ln t+t(\ln t)')dt=(\ln t+\frac{t}{t})dt$$ $$du=(\ln t+1)dt$$
- For $t=2$, we have $u=2\ln2$
- For $t=4$, we have $u=4\ln4$
Therefore, $$A=\int^{4\ln4}_{2\ln2}udu=\frac{u^2}{2}\Big]^{4\ln4}_{2\ln2}$$ $$A=\frac{1}{2}\Big(16(\ln4)^2-4(\ln2)^2\Big)$$ $$A=8(\ln4)^2-2(\ln2)^2$$ $$A=8(\ln2^2)^2-2(\ln2)^2$$ $$A=8(2\ln2)^2-2(\ln2)^2$$ $$A=32(\ln2)^2-2(\ln2)^2=30(\ln2)^2$$
