# Integration by Parts

## What is it

When there are 2 functions(composite function) in an integration operator, we can use the method, Integration by Parts.

It is used for combination of functions in an Integral. For examples, Polynomial functions($$x,\ x^2,\ x^3, ・・・)$$, Exponential functions$$(e^x,\ 2^x, ・・・)$$, Trigonometric functions$$(\sin{x},\ \cos{x}, ・・・)$$, Logarithmic functions$$(\log{x},\ \ln{x})$$ and so on.

This is one of most important formula for EJU Math. So let’s memorize it.

## How to use

Example 1

$$\int x e^x dx$$

This is combination of a polynomial($$x$$) and Exponential function ($$e^x$$).

We can use this formula to solve it.

$$\int u v’ dx= u v -\int u’ v’ dx$$

$$v’$$ indicates the differentiation of $$v$$. And please remember that I avoid writing like $$u(x)$$ because $$u$$ and $$v$$ are obviously functions of $$x$$.

We set

$$\color{blue}{u=x},\ v’=e^x$$

Now we have to find $$u’$$ and $$v$$

$$\color{red}{u’=1},\ \color{purple}{v=e^x}$$

Then we substitute those values to appropriate places.

Since the formula is

$$\int u v’ dx= \color{blue}{u} \color{purple}{v} -\int \color{red}{u’} \color{purple}{v} \ dx$$

$$\int u v’ dx= (\color{blue}{x}) (\color{purple}{e^x}) -\int (\color{red}{1}) (\color{purple}{e^x}) dx$$

$$=x e^x-e^x+C$$

$$=e^x(x-1)+C$$

## Deviation for Integration by Parts

Let’s differentiate a composite function$$uv$$

Since it is a composite function, we need to use product rule for the differentiation.

$$(uv)’=uv’+u’v$$

Then we integrate for both sides

$$\int (uv)’ dx= \int uv’ dx + \int u’v \ dx$$

If we isolate the second term in the right side, $$\int uv’ dx$$

$$\int uv’ dx = \int (uv)’ dx – \int u’v \ dx$$

Since $$\int (uv)’ dx = uv$$, we eventually have

$$\int uv’ dx = uv – \int u’v\ dx$$

This is the formula for Integration by Parts.

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