evaluate $int_{1/3}^{5}14xln(3x)dx$.\n$int_{1/3}^{5}14xln(3x)dx=square$ (type an exact answer.)

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \ln(3x)$ and $\mathrm{d}v=14x\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v = 7x^{2}$.

Step2: Apply the formula

$\int 14x\ln(3x)\mathrm{d}x=7x^{2}\ln(3x)-\int 7x^{2}\cdot\frac{1}{x}\mathrm{d}x=7x^{2}\ln(3x)-\int 7x\mathrm{d}x$.

Step3: Integrate the remaining part

$\int 7x\mathrm{d}x=\frac{7}{2}x^{2}+C$. So, $\int 14x\ln(3x)\mathrm{d}x=7x^{2}\ln(3x)-\frac{7}{2}x^{2}+C$.

Step4: Evaluate the definite integral

$\left[7x^{2}\ln(3x)-\frac{7}{2}x^{2}\right]_{\frac{1}{3}}^{5}=7\times5^{2}\ln(15)-\frac{7}{2}\times5^{2}-\left(7\times\left(\frac{1}{3}\right)^{2}\ln(1)-\frac{7}{2}\times\left(\frac{1}{3}\right)^{2}\right)$. Since $\ln(1) = 0$, we have: [ \begin{align*} &7\times25\ln(15)-\frac{175}{2}+\frac{7}{18}\ =&175\ln(15)-\frac{1575 - 7}{18}\ =&175\ln(15)-\frac{1568}{18}\ =&175\ln(15)-\frac{784}{9} \end{align*} ]

Answer:

$175\ln(15)-\frac{784}{9}$