evaluate ∫₂⁷ t³ ln(3t) dt.

Explanation:

Step1: Use integration - by - parts

The formula for integration by parts is $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$. Let $u = \ln(3t)$ and $dv=t^{3}dt$. Then $du=\frac{1}{t}dt$ and $v=\frac{t^{4}}{4}$.

Step2: Apply the integration - by - parts formula

$\int_{2}^{7}t^{3}\ln(3t)dt=\left[\frac{t^{4}}{4}\ln(3t)\right]{2}^{7}-\int{2}^{7}\frac{t^{4}}{4}\cdot\frac{1}{t}dt$. First, evaluate $\left[\frac{t^{4}}{4}\ln(3t)\right]{2}^{7}=\frac{7^{4}}{4}\ln(21)-\frac{2^{4}}{4}\ln(6)=\frac{2401}{4}\ln(21) - 4\ln(6)$. Second, simplify the second integral: $\int{2}^{7}\frac{t^{4}}{4}\cdot\frac{1}{t}dt=\frac{1}{4}\int_{2}^{7}t^{3}dt$.

Step3: Evaluate the remaining integral

$\frac{1}{4}\int_{2}^{7}t^{3}dt=\frac{1}{4}\left[\frac{t^{4}}{4}\right]_{2}^{7}=\frac{1}{16}(7^{4}-2^{4})=\frac{1}{16}(2401 - 16)=\frac{2385}{16}$.

Step4: Combine the results

$\int_{2}^{7}t^{3}\ln(3t)dt=\frac{2401}{4}\ln(21)-4\ln(6)-\frac{2385}{16}$.

Answer:

$\frac{2401}{4}\ln(21)-4\ln(6)-\frac{2385}{16}$