- #1
Dustinsfl
- 2,281
- 5
$$
\int (\ln x)^pdx, \quad p > 0
$$
How do I go about integrating this?
\int (\ln x)^pdx, \quad p > 0
$$
How do I go about integrating this?
Ackbach said:Is $p$ an integer? If so, then you could do $u=\ln(x)$, which leads to
$$\int u^{p}e^{u}\,du,$$
which succumbs to by-parts as many times as you need. I'd recommend tabular integration in that case. If $p$ can be real, you get something a bit more nasty, with the incomplete Gamma function in there. I don't know how you get that result. I'd have to study a bit.
The purpose of integrating Ln $(x)^p$ is to find the antiderivative of a function that includes the natural logarithm raised to a power. This allows us to evaluate the area under the curve of the function, which is useful in many mathematical and scientific applications.
The first step in integrating Ln $(x)^p$ is to use the power rule to bring the exponent down in front of the natural logarithm. Then, we can use the formula for integrating Ln $x$ to solve the integral. Finally, we substitute the original value of $x$ back into the equation to find the definite integral.
The power rule for integrating Ln $(x)^p$ states that the integral of Ln $(x)^p$ is equal to $x^p \ln x - \frac{x^p}{p+1} + C$.
For example, if we want to integrate Ln $(x)^3$, we first use the power rule to bring the exponent down: $3 \cdot$ Ln $x$. Then, we use the formula for integrating Ln $x$: $x \ln x - x + C$. Finally, we substitute the original value of $x$ back into the equation to find the definite integral.
Yes, there are two special cases when integrating Ln $(x)^p$: when $p = 0$ and when $p = 1$. When $p = 0$, the integral simplifies to just $x + C$. When $p = 1$, the integral becomes $x \ln x - x + C$. It is important to be aware of these special cases when integrating Ln $(x)^p$.