Question

use integration by parts to prove the reduction formula:\{\int {{{\left( {lnx} \right)}^n}} dx = x{\left( {lnx} \right)^n} - n\int {{{\left( {lnx} \right)}^{n - 1}}} dx\)

Short answer

We can prove the following reduction formula by using integration by parts:

\{\int {{{\left( {\ln x} \right)}^n}dx = x{{\left( {\ln x} \right)}^n} - n} \int {{{\left( {\ln x} \right)}^{n - 1}}dx} \)

Step-by-step solution

Use the integration of parts

\{\begin{aligned}{l}uv = \int {vdu} + \int {udv} \\\int {udv = uv - \int {vdu\,\,\,\,\,\,\,\,\,\,........\left( 1 \right)} } \end{aligned}\)

Let \{\begin{aligned}{l}u = {\left( {\ln x} \right)^n}\\du = \frac{{n{{\left( {\ln x} \right)}^{n - 1}}}}{x}dx\\dv = dx\\v = x\end{aligned}\)

Substitute \{u,v,du,dv\) in the above equation

\{\begin{aligned}{l}\int {udv = uv - \int {vdu} } \\\int {{{\left( {\ln x} \right)}^n}dx = {{\left( {\ln x} \right)}^n}x - } \int {x\frac{{n{{\left( {\ln x} \right)}^{n - 1}}}}{x}dx} \\\int {{{\left( {\ln x} \right)}^n}dx = {{\left( {\ln x} \right)}^n}x - } n\int {{{\left( {\ln x} \right)}^{n - 1}}dx} \end{aligned}\)

Hence, we can prove that for the reduction formula by using integration by parts:

\{\int {{{\left( {\ln x} \right)}^n}dx = x{{\left( {\ln x} \right)}^n} - n} \int {{{\left( {\ln x} \right)}^{n - 1}}dx} \).