evaluate the integral using integration by parts.\n int_{1}^{e^{5}}x^{3}ln(x)dx \n int_{1}^{e^{5}}x^{3}ln(x)d…

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 $\mathrm{d}v=x^{3}\mathrm{d}x$.

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

Differentiate $u=\ln(x)$ with respect to $x$: $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$. Integrate $\mathrm{d}v = x^{3}\mathrm{d}x$ with respect to $x$: $v=\frac{x^{4}}{4}$.

Step3: Apply integration - by - parts formula

$\int_{1}^{e^{5}}x^{3}\ln(x)\mathrm{d}x=\left[\frac{x^{4}}{4}\ln(x)\right]{1}^{e^{5}}-\int{1}^{e^{5}}\frac{x^{4}}{4}\cdot\frac{1}{x}\mathrm{d}x$.

Step4: Evaluate $\left[\frac{x^{4}}{4}\ln(x)\right]_{1}^{e^{5}}$

$\left[\frac{x^{4}}{4}\ln(x)\right]_{1}^{e^{5}}=\frac{(e^{5})^{4}}{4}\ln(e^{5})-\frac{1^{4}}{4}\ln(1)$. Since $\ln(e^{5}) = 5$ and $\ln(1)=0$, we have $\frac{e^{20}}{4}\cdot5-0=\frac{5e^{20}}{4}$.

Step5: Evaluate $\int_{1}^{e^{5}}\frac{x^{3}}{4}\mathrm{d}x$

$\int_{1}^{e^{5}}\frac{x^{3}}{4}\mathrm{d}x=\frac{1}{4}\cdot\left[\frac{x^{4}}{4}\right]_{1}^{e^{5}}=\frac{1}{16}\left[(e^{5})^{4}-1^{4}\right]=\frac{e^{20}-1}{16}$.

Step6: Calculate the final result

$\int_{1}^{e^{5}}x^{3}\ln(x)\mathrm{d}x=\frac{5e^{20}}{4}-\frac{e^{20}-1}{16}$. Find a common denominator: $\frac{20e^{20}-(e^{20}-1)}{16}=\frac{20e^{20}-e^{20}+1}{16}=\frac{19e^{20}+1}{16}$.

Answer:

$\frac{19e^{20}+1}{16}$