The integration of the UV formula, often referred to as integration by parts, is a technique used in calculus. It is based on the product rule for differentiation and is used to integrate the product of two functions. The formula is expressed as:

$$ \int u \, dv = uv – \int v \, du $$

**How to use it**

(1) Split the integrand into two parts, $ u $ and $ dv $, such that $ u $ is a function of $ x $ and $ dv $ is a differential of another function of $ x $. The choice of $ u $ and $ dv $ is crucial and can simplify the problem significantly. A common strategy is to choose $ u $ as the part that becomes simpler when differentiated, and $ dv $ as the part that is easy to integrate.

We can decide the u function using the word ILATE

I -> Inverse trigonometric functions

L -> Logarithmic functions

A-> Algebraic functions

T -> trigonometric functions

E -> Exponential functions

(2) Differentiate $ u $ with respect to $ x $ to find $ du $.

(3) Integrate $ dv $ with respect to $ x $ to find $ v $.

(4) Substitute $ u $, $ du $, and $ v $ into the integration by parts formula and simplify.

**Examples**

**Question 1**

$\int x \cdot e^x \; dx$

**Solution**

Let $ u = x $. This is a good choice because its derivative is simpler.

Let $ dv = e^x dx $. This is chosen because the exponential function is easy to integrate.

Then

$ du = dx $.

and

$ v = \int e^x dx = e^x $.

Therefore

$ \int x \cdot e^x dx = x \cdot e^x – \int e^x \cdot dx \\

= x \cdot e^x – e^x + C $

**Example 2**

$ \int x^2 e^x \, dx $.

**Solution**

$ u = x^2 $ (since its derivative is simpler).

$ dv = e^x dx $ (easy to integrate).

Then

$ du = 2x dx $.

and

$ v = \int e^x dx = e^x $.

Therefore

$ \int x^2 e^x dx = x^2 e^x – \int 2x e^x dx $.

Now, we need to integrate $ \int 2x e^x dx $ by parts again.

For $ \int 2x e^x dx $:

Choose $ u = 2x $ and $ dv = e^x dx $.

Then, $ du = 2 dx $ and $ v = e^x $.

So, $ \int 2x e^x dx = 2x e^x – \int 2e^x dx = 2x e^x – 2e^x $.

Substituting this back into our original equation:

$ x^2 e^x – (2x e^x – 2e^x) $.

So, the integral is $ x^2 e^x – 2x e^x + 2e^x + C $.

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