# Integrals of Gradshteyn and Ryzhik: 2.723 – Combinations of Logarithms and Algebraic Functions

The strategy for all three integrals is to use a different integrand, parameterized, differentiate under the integral, then apply a limit to recover the desired integral.

## 2.723.1

The integral to be evaluated is

I = \int x^{n}\ln(x) dx

Let

I(a) = \int x^{a-1} dx = \frac{x^a}{a}

so that
\begin{align}
\frac{\partial}{\partial a}I(a) &= \int x^{a-1}\ln(x) dx \\
&= \frac{\partial}{\partial a} \frac{x^a}{a} = x^{a} \left[\frac{\ln(x)}{a}\,- \frac{1}{a^2} \right]
\end{align}

\lim_{a \to n+1} \frac{\partial}{\partial a}I(a) \Rightarrow
I = \bbox[5px,border:2px solid blue] {\int x^{n}\ln(x) dx = x^{n+1} \left[\frac{\ln(x)}{n+1}\,- \frac{1}{(n+1)^2} \right]}

## 2.723.2

The integral to be evaluated is

I = \int x^{n}\ln^{2}(x) dx

\begin{align}
\frac{{\partial}^2}{\partial a^2}I(a) &= \int x^{a-1}\ln^{2}(x) dx \\
&= \frac{\partial}{\partial a} \left[\ln(x) \frac{x^a}{a}\,- \frac{x^a}{a^2} \right]
= x^{a} \left[\frac{\ln^{2}(x)}{a}\, – \frac{2\ln(x)}{a^2} + \frac{2}{a^3} \right]
\end{align}

\lim_{a \to n+1} \frac{{\partial}^2}{\partial a^2}I(a) \Rightarrow
I = \bbox[5px,border:2px solid blue] {\int x^{n}\ln^{2}(x) dx
= x^{n+1} \left[\frac{\ln^{2}(x)}{n+1}\, – \frac{2\ln(x)}{(n+1)^2} + \frac{2}{(n+1)^3} \right]}

## 2.723.3

The integral to be evaluated is

I = \int x^{n}\ln^{3}(x) dx

\begin{align}
\frac{{\partial}^3}{\partial a^3}I(a) &= \int x^{a-1}\ln^{3}(x) dx \\
&= \frac{\partial}{\partial a} \left( x^{a} \left[\frac{\ln^{2}(x)}{a}\, – \frac{2\ln(x)}{a^2} + \frac{2}{a^3} \right] \right) \\
&= x^{a} \left[\frac{\ln^{3}(x)}{a} \,- \frac{3\ln^{2}(x)}{a^2} + \frac{6\ln(x)}{a^3} \,- \frac{6}{a^4} \right]
\end{align}

\lim_{a \to n+1} \frac{{\partial}^2}{\partial a^3}I(a) \Rightarrow

I = \bbox[5px,border:2px solid blue] {\int x^{n}\ln^{2}(x) dx
= x^{n+1} \left[\frac{\ln^{3}(x)}{n+1} \,- \frac{3\ln^{2}(x)}{(n+1)^2} + \frac{6\ln(x)}{(n+1)^3} \,- \frac{6}{(n+1)^4} \right]}