evaluate the integral using integration by parts.\n int_{1}^{e^{5}} x^{2}ln(x)dx \n int_{1}^{e^{5}}…

Explanation:

Step1: Recall 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 $\mathrm{d}v=x^{2}\mathrm{d}x$.

Step2: Find $\mathrm{d}u$ and $v$

Differentiate $u=\ln(x)$ to get $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$. Integrate $\mathrm{d}v = x^{2}\mathrm{d}x$ to get $v=\frac{1}{3}x^{3}$.

Step3: Apply the integration - by - parts formula

$\int_{1}^{e^{5}}x^{2}\ln(x)\mathrm{d}x=\left[\frac{1}{3}x^{3}\ln(x)\right]{1}^{e^{5}}-\int{1}^{e^{5}}\frac{1}{3}x^{3}\cdot\frac{1}{x}\mathrm{d}x$. First, evaluate $\left[\frac{1}{3}x^{3}\ln(x)\right]{1}^{e^{5}}$. When $x = e^{5}$, $\frac{1}{3}(e^{5})^{3}\ln(e^{5})=\frac{1}{3}e^{15}\cdot5=\frac{5}{3}e^{15}$. When $x = 1$, $\frac{1}{3}(1)^{3}\ln(1)=0$. Second, simplify the second integral: $\int{1}^{e^{5}}\frac{1}{3}x^{3}\cdot\frac{1}{x}\mathrm{d}x=\frac{1}{3}\int_{1}^{e^{5}}x^{2}\mathrm{d}x$.

Step4: Evaluate the remaining integral

$\frac{1}{3}\int_{1}^{e^{5}}x^{2}\mathrm{d}x=\frac{1}{3}\left[\frac{1}{3}x^{3}\right]{1}^{e^{5}}=\frac{1}{9}(x^{3})|{1}^{e^{5}}=\frac{1}{9}(e^{15}-1)$.

Step5: Calculate the final result

$\int_{1}^{e^{5}}x^{2}\ln(x)\mathrm{d}x=\frac{5}{3}e^{15}-\frac{1}{9}(e^{15}-1)=\frac{15e^{15}-e^{15}+1}{9}=\frac{14e^{15}+1}{9}$.

Answer:

$\frac{14e^{15}+1}{9}$