use integration by parts to evaluate the integral: $int_{1}^{4}\frac{ln(x)}{x^{3}}dx = $ report your answer…

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 $\mathrm{d}v=\frac{1}{x^{3}}\mathrm{d}x$.

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 $\mathrm{d}v=\frac{1}{x^{3}}\mathrm{d}x=x^{-3}\mathrm{d}x$: $v=\int x^{-3}\mathrm{d}x=-\frac{1}{2x^{2}}$.

Step3: Apply the integration - by - parts formula

$\int_{1}^{4}\frac{\ln(x)}{x^{3}}\mathrm{d}x=\left[-\frac{\ln(x)}{2x^{2}}\right]{1}^{4}-\int{1}^{4}\left(-\frac{1}{2x^{2}}\right)\cdot\frac{1}{x}\mathrm{d}x$. First, evaluate $\left[-\frac{\ln(x)}{2x^{2}}\right]{1}^{4}=-\frac{\ln(4)}{2\times4^{2}}+\frac{\ln(1)}{2\times1^{2}}=-\frac{\ln(4)}{32}+0=-\frac{\ln(4)}{32}$. Second, simplify the second integral: $\int{1}^{4}\left(-\frac{1}{2x^{2}}\right)\cdot\frac{1}{x}\mathrm{d}x=-\frac{1}{2}\int_{1}^{4}x^{-3}\mathrm{d}x$.

Step4: Evaluate the remaining integral

$-\frac{1}{2}\int_{1}^{4}x^{-3}\mathrm{d}x=-\frac{1}{2}\left[\frac{x^{-2}}{-2}\right]{1}^{4}=\frac{1}{4}\left[\frac{1}{x^{2}}\right]{1}^{4}=\frac{1}{4}\left(\frac{1}{16}-1\right)=\frac{1}{4}\times\left(-\frac{15}{16}\right)=-\frac{15}{64}$.

Step5: Calculate the final result

$\int_{1}^{4}\frac{\ln(x)}{x^{3}}\mathrm{d}x=-\frac{\ln(4)}{32}+\frac{15}{64}$. Using $\ln(4)\approx1.386294$, we have $-\frac{1.386294}{32}+\frac{15}{64}=- 0.043322+0.234375 = 0.191053$.

Answer:

$0.191053$