Question

consider the integral $\int x^{5}\ln(x)dx$: applying the integration by parts technique, let $u =$ and $dv = dx$ then $uv-\int vdu=$ $-\int dx$

Explanation:

Step1: Choose \(u\) and \(dv\)

According to the integration - by - parts formula \(\int u\;dv=uv-\int v\;du\), we choose \(u = \ln(x)\) (because its derivative is simpler) and \(dv=x^{5}dx\) (because we can integrate \(x^{n}\) easily using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C,n
eq - 1\)).

Step2: Find \(du\) and \(v\)

Differentiate \(u=\ln(x)\) using the formula \((\ln(x))^\prime=\frac{1}{x}\), so \(du=\frac{1}{x}dx\).
Integrate \(dv = x^{5}dx\) using the power rule \(\int x^{n}dx=\frac{x^{n+1}}{n + 1}+C\). Here \(n = 5\), so \(v=\int x^{5}dx=\frac{x^{6}}{6}\).

Step3: Apply the integration - by - parts formula

Substitute \(u\), \(v\), \(du\) into the formula \(uv-\int v\;du\).
We get \(uv-\int v\;du=\frac{x^{6}}{6}\ln(x)-\int\frac{x^{6}}{6}\cdot\frac{1}{x}dx\).
Simplify the integrand \(\frac{x^{6}}{6}\cdot\frac{1}{x}=\frac{x^{5}}{6}\).

Answer:

\(u=\ln(x)\), \(dv = x^{5}dx\), \(uv-\int v\;du=\frac{x^{6}}{6}\ln(x)-\int\frac{x^{5}}{6}dx\)