Calculate $\int x^m \ln(x)\,dx,\,\,\, m \in \mathbb{Z}$

Question

Calculate

$$\int x^m \ln(x)\,dx,\,\,\, m \in \mathbb{Z}$$

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My attempt:

First suppose $m\ne-1$

$$\int x^m \ln(x)\,dx=\left[\frac{1}{(m+1)}x^{m+1}\ln(x)\right]-\int \frac{1}{(m+1)}x^{m+1}x^{-1}\,dx$$

$$=\left[\frac{1}{(m+1)}x^{m+1}\ln(x)\right]-\int \frac{1}{(m+1)}x^mdx=\left[\frac{1}{(m+1)}x^{m+1}\ln(x)\right]-\left[\frac{1}{(m+1)^2}x^{m+1}\right]$$

$$=\left[\frac{1}{(m+1)}x^{m+1}\ln(x)-\frac{1}{(m+1)^2}x^{m+1}\right]$$

Now suppose $m=-1$

$$\int x^{-1} \ln(x)\,dx=\left[(\ln(x))^2\right]-\int \ln(x)x^{-1}\,dx$$

$$\Longleftrightarrow 2\int x^{-1} \ln(x)\,dx=\left[(\ln(x))^2\right]$$

$$\Longrightarrow \int x^{m} \ln(x) \, dx=\frac{1}{2} \left[(\ln(x))^2\right], m=-1$$

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Hey it would be great if someone could check my attempt to solve the task :) thank you

Answer 1

You solution looks correct. I obtain the same result using an alternative approach.

Substitute $ x = e^y $, the integral will transform to,

$$I = \int ye^{(m+1)y} dy$$

For $m \neq -1$, by using integration by parts we get,

$$I = \frac{ye^{(m+1)y}}{m+1} - \frac{e^{(m+1)y}}{(m+1)^2} $$

Replacing $y$ gives,

$$I = \frac{\ln(x)}{m+1}x^{(m+1)} - \frac{x^{(m+1)}}{(m+1)^2} $$

For $ m = -1$,

$$ I = \int y dy = \frac{1}{2}y^2$$

Replacing $y$ gives,

$$ I = \frac{1}{2} [\ln(x)]^2 $$

Answer 2

Let $x=t^{1/(m+1)}$. By magic, the factor $x^m$ cancels away.

$$\int x^m\log(x)\,dx=\frac1{(m+1)^2}\int \log t\,dt=\frac{t(\log t-1)}{(m+1)^2}=\frac{x^{m+1}(\log x^{m+1}-1)}{(m+1)^2}.$$

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For $m=-1$,

$$\int\frac{\log t}t\,dt=\int\log t\,d\log t=\frac{\log^2t}2.$$