Formula for Integration by Parts
$$\begin{aligned}&\int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)dx\\\\&\text{or}\quad\int ud\nu=u\nu-\int vdu\end{aligned}$$
How to pick \(f(x)\) and \(g'(x)\). i.e., \(u\) and \(dv\)?
Here is a suggestion!
Remember the word LATE:
L - logarithmic ex. \(ln(x)\)
A - algebraic ex. \(x^{2}+x\)
T - trigonometric ex. \(sin(x)\)
E - exponential ex. \(e^{x}\)
The first expression that appears in the word LATE in the integrand will be \(u\), and the rest will be \(dv\).
Take the derivative of \(u\) and integrate \(dv\). Then, use the formula for integration by parts.
Example: Find \(\int xe^{2x}dx.\)
Solution: According to LATE, \(u=x\) and \(d\nu=e^{2x}dx.\)
\(\begin{aligned}u&=x\\du&=dx\end{aligned}\) and \(\begin{aligned}\int d\nu&=\int e^{2x}dx\\\nu&=\frac{e^{2x}}2\end{aligned}\)
\(\begin{aligned}\int xe^{2x}dx&=(x)\bigg(\frac{e^{2x}}{2}\bigg)-\int\frac{e^{2x}}{2}dx=\frac{xe^{2x}}{2}-\frac{1}{2}\bigg(\frac{e^{2x}}{2}\bigg)+C\\&=\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}+C=\frac{e^{2x}}{2}\bigg(x-\frac{1}{2}\bigg)+C\end{aligned}\)
Definite Integration by Parts
The integration by parts formula can be applied to definite integrals by noting that
$$\int_{b}^{a}udv=uv\Big|_{a}^{b}-\int_{b}^{a}vdu.$$
Computing a Definite Integral by Integration by Parts
Step 1. Solve the integral as an indefinite integral.
Step 2. Evaluate it over the interval of integration.
Example: Find \(\int_1^e\frac{\ln x}{x^2}dx.\)
Solution:
Step 1. According to LATE,
\(u=lnx\) and \(d\nu={\frac{1}{x^{2}}}dx=x^{-2}dx.\)
\(\begin{aligned}u&=\ln x\\du&=\frac1xdx\end{aligned}\) and \(\begin{aligned}\int d\nu&=\int\frac{1}{x^2}dx=\int x^{-2}dx\\\nu&=\frac{-1}{x}\end{aligned}.\)
\(\begin{gathered}
\int\frac{\ln x}{x^{2}}dx =(\ln x)\biggl(\frac{-1}x\biggr)-\int\biggl(\frac{-1}x\biggr)\biggl(\frac1xdx\biggr) \\
=-\frac{\ln x}{x}-\int x^{-2}dx \\
=-\frac{\ln x}{x}-\frac{1}{x}+C
\end{gathered}\)
Step 2.
\(\begin{aligned}
\int_{1}^{e}{\frac{\ln x}{x^{2}}}dx& =-\frac{\ln x}{x}-\frac{1}{x}\biggr|_{1}^{e} \\
&=\left(-\frac{\ln e}{e}-\frac{1}{e}\right)-\left(-\frac{\ln1}{1}-\frac{1}{1}\right) \\
&=-\frac2e+1
\end{aligned}\)