# What is integration by Parts?

Oct 2, 2014

When you come across a function that cannot be easily antidifferentiated, but some part of it can be easily antidifferentiated, we need to use integration by parts.

The rule for integration by parts is:

${\int}_{a}^{b} f \left(x\right) \cdot g ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right) \cdot g \left(x\right)\right]}_{a}^{b} - {\int}_{a}^{b} f ' \left(x\right) \cdot g \left(x\right) \mathrm{dx}$

Say we wish to integrate the function $h \left(x\right) = {e}^{x} \cdot x$
We can split this into two functions:

$\text{Let "f(x)=x" and } g \left(x\right) = {e}^{x}$
$\implies f ' \left(x\right) = 1 \mathmr{and} g ' \left(x\right) = {e}^{x}$
$\implies f \left(x\right) \cdot g ' \left(x\right) = {e}^{x} \cdot x$

This is precisely what we need to integrate, so all we do is plug f(x), f'(x), g(x) and g'(x) into the above formula:

${\int}_{a}^{b} x \cdot {e}^{x} \mathrm{dx} = {\left[x \cdot {e}^{x}\right]}_{a}^{b} - {\int}_{a}^{b} 1 \cdot {e}^{x} \mathrm{dx}$

Note that the right-hand side of the equation is much "nicer" to work with than before.

$\implies {\int}_{a}^{b} x \cdot {e}^{x} \mathrm{dx} = b {e}^{b} - a {e}^{a} - \left({e}^{b} - {e}^{a}\right)$

(I have skipped a couple of steps)

The trick to integrating by parts is to:

1. Select f(x) so that you can easily find f'(x). If you can't find f'(x) easily, then try again.
2. Select g'(x) so that you can easily find g(x). Again, if g'(x) has no antiderivative, or it's not easy to find an antiderivative, try something else.
3. Select f(x) and g'(x) so that f'(x)g(x) can be easily integrated. If you come up with an integral that needs to be integrated by parts again and again, there's probably an easier way of doing it.
Jul 29, 2015

When one has to integrate a product of two functions, integration by parts is useful.

If $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

then $\int f \left(x\right) \mathrm{dx} = g \left(x\right) \int h \left(x\right) \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} g \left(x\right) \cdot \int h \left(x\right) \mathrm{dx}\right) \mathrm{dx}$
This is called integration by parts.

#### Explanation:

The integral of the product of two functions may be verbally given as,

"First function into integral of the second minus integral of the derivative of the first into integral of the second." Which is nothing but, integration by parts.

Now, one thing that must be noted is that, the correct choice of first function and second function can either make or break a problem.
The correct choice can vastly simplify and an incorrect one can put you in a lot of trouble.

For instance,

$\int x \cdot {e}^{x} \mathrm{dx}$ is the integral we need to evaluate.
If we use ${e}^{x}$ as the first function and $x$ as the second and integrate by parts,

$\int x \cdot {e}^{x} \cdot \mathrm{dx} = {e}^{x} \int x \cdot \mathrm{dx} - \int \left({e}^{x} \int x \cdot \mathrm{dx}\right) \cdot \mathrm{dx}$
$= {e}^{x} \cdot {x}^{2} / 2 - \int {e}^{x} \cdot {x}^{2} / 2 \cdot \mathrm{dx} + C$

If we apply integration by parts to the second term, we again get a term with a ${x}^{3}$ and so on.

This, not only complicates the problem but, spells disaster.

But, if we had chosen $x$ to be the first and ${e}^{x}$ to be the second, the integral would have been very simply to evaluate.

$\int x \cdot {e}^{x} \cdot \mathrm{dx} = x \int {e}^{x} \cdot \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} x \int {e}^{x} \cdot \mathrm{dx}\right) \cdot \mathrm{dx}$
$= {e}^{x} \cdot x - {e}^{x} + C$