4. Integration by Parts
Recall: \(\displaystyle \int u\,dv=u\,v-\int v\,du\) where \(du=\dfrac{du}{dx}\,dx\) and \(dv=\dfrac{dv}{dx}\,dx\)
b1. Choosing the 'Parts'
b. Log Examples
Compute \(\displaystyle \int x^4\ln x\,dx\).
Normally we would take the polynomial \(x^4\) as \(u\). However, since it is hard to integrate \(\ln x\), we take \[\begin{array}{ll} u=\ln x & dv=x^4\,dx \\ du=\dfrac{1}{x}\,dx \quad & v=\dfrac{x^5}{5} \end{array}\] This is the “L” in LAPTE. Then, the integration by parts gives: \[\begin{aligned} \int x^4\ln x\,dx &=\dfrac{x^5}{5}\ln x-\int \dfrac{x^5}{5}\dfrac{1}{x}\,dx \\ &=\dfrac{x^5}{5}\ln x-\int \dfrac{x^4}{5}\,dx \\ &=\dfrac{x^5}{5}\ln x-\dfrac{x^5}{25}+C \end{aligned}\]
We check by differentiating (using the Product Rule): If \(f(x)=\dfrac{x^5}{5}\ln x-\dfrac{x^5}{25}\), then \[ f'(x)=x^4\ln x+\dfrac{x^5}{5}\dfrac{1}{x}-\dfrac{x^4}{5} =x^4\ln x \] which is the integrand we started with.
In this example, the general rule about selecting for \(dv\) a function easy to integrate supersedes the rule about taking \(u\) as the polynomial.
Compute the integral \(\displaystyle \int x^9\ln x\,dx\).
\(\displaystyle \int x^9\ln x\,dx =\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}+C\)
We let: \[\begin{array}{ll} u=\ln x & dv=x^9\,dx\\ du=\dfrac{1}{x}\,dx \quad & v=\dfrac{1}{10}x^{10} \end{array}\] So we get: \[\begin{aligned} \int x^9\ln x\,dx &=\dfrac{1}{10}x^{10}\ln x-\int \dfrac{1}{10}x^9\,dx \\ &=\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}+C \end{aligned}\]
We check using the Product Rule: \[\begin{aligned} \dfrac{d}{dx}&\left(\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}\right) \\ &=x^9\ln x+\dfrac{1}{10}x^{10}\dfrac{1}{x}-\dfrac{1}{10}x^9 =x^9\ln x \end{aligned}\]