Integration By Parts Formula
Do you remember that we were using the product rule to differentiate the product of two functions in differentiation? In the same way, the integration by parts formula is used to find the integral of the product of two different types of functions such as logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions. Thus, it can be called as product rule of integration (though there is no such name for this). Let us learn the integration by parts formula along with a few solved examples.
What Is Integration By Parts Formula?
In the product rule of differentiation where we differentiate a product uv, u and v can be chosen in any order. But while using the integration by parts formula, for choosing the first function u, we have to see which of the following function comes first in the following order and then assume it as u.
 Logarithmic (L)
 Inverse trigonometric (I)
 Algebraic (A)
 Trigonometric (T)
 Exponential (E)
This can be remembered using the rule LIATE. Note that this order can be ILATE as well. For example, if we have to find \(\int x \ln x\) (where x is an algebraic function and ln is a logarithmic function), we will choose \(\ln x\) to be u as in LIATE, the logarithmic function appears before the algebraic function. The integration by parts formula is defined in two ways. We can use either of them to integrate the product of two functions.
\( \begin{align} \int uv \, dx &= u \int v \, dx  \int \left( u' \, \int v\, dx \right) \, dx\\[0.2cm] \int u\, dv &= uv  \int v \,du \end{align}\)
Solved Examples Using Integration By Parts Formula

Example 1:
Evaluate the integral \(\int x \ln x \, dx\) using integration by parts.
Solution:
First Method:
Using LIATE, u = ln x and v = x.
Using one of the formulas of integration by parts,
\( \begin{align} \int uv \, dx &= u \int v \, dx  \int \left( u' \, \int v\, dx \right) \, dx\\[0.2cm] \int x \ln x \, dx &= \ln x \int x \, dx  \int \left( \dfrac{d}{dx}(\ln x) \int x \, dx \right) \, dx\\[0.2cm] &= \ln x \cdot \dfrac{x^2}{2} \int \dfrac{1}{x} \cdot \dfrac{x^2}{2} dx\\[0.2cm] &= \dfrac{x^2 \ln x}{2}  \dfrac{1}{2} \int x \, dx \\ &= \dfrac{x^2 \ln x}{2}  \dfrac{x^2}{4}+C \end{align}\)
Second Method:
Using LIATE, u = ln x and dv = x dx.
Then \(du = \dfrac{1}{x} \,dx\) and \(v = \int x \, dx = \dfrac{x^2}{2}\)
Using one of the formulas of integration by parts,
\( \begin{align} \int u\, dv &= uv\int v\, du \\[0.2cm] \int \ln x\, (x \, dx) &= \ln x \left( \dfrac{x^2}{2}\right)  \int \dfrac{x^2}{2} \cdot \dfrac{1}{x}dx\\[0.2cm] &= \dfrac{x^2 \ln x}{2}  \dfrac{1}{2} \int x \, dx \\ &= \dfrac{x^2 \ln x}{2}  \dfrac{x^2}{4}+C \end{align}\)Answer: By both the methods, \(\int x \ln x \, dx = \dfrac{x^2 \ln x}{2}  \dfrac{x^2}{4}+C \)

Example 2
Evaluate the integral \(\int x^2 e^x \, dx\).
Solution:
Using LIATE, u = x^{2} and dv = e^{x} dx.
Then, du = 2x dx, \(v = \int e^x dx = e^x\).
Using one of the LIATE formulas,
\( \begin{align} \int u\, dv &= uv\int v\, du \\[0.2cm] \int x^2 e^x dx &= x^2e^x \int e^x (2x \, dx)\\[0.2cm] &= x^2 e^x  2 \int x e^x dx \end{align}\)
Applying integration by parts formula again to evaluate \(\int x e^x dx\),
\( \begin{align} \int x^2e^x dx &= x^2e^x 2 \left( xe^x  \int e^x dx \right) \\[0.2cm] &= x^2e^x  2xe^x + 2e^x +C \end{align} \)
Answer: \(\int x^2 e^x \, dx = x^2e^x  2xe^x + 2e^x +C\)