integration by parts -integration techniqueslets

Question

Integration by Parts -Integration Techniques

Let's start off along with this section with a couple of integrals that we should previously be able to do to get us started. Firstly let's take a look at the following.

∫ e^x dx = e^x + c

Thus, that was simple enough.  Now, let's take a look at,

∫ xe^x2 dx

To do this integral we'll make use of the following substitution.

U = x²   du=2xdx => xdx = ½ du

∫ xe^x2 dx = ½ ∫ e^u du ½ e^u + c=1/2 e^x2 + c

Once Again, simple enough to do offer you remember how to do substitutions.  By the way ensure that you can do these types of substitutions quickly and easily.  From this point on we are going to be doing these types of substitutions in our head.  If you have to prevent and write these out with each problem you will find out that it will take you considerably longer to do these problems.

Now, let's look at the integral that we really wish to do.

∫ xe^6x dx

If we just had an x by itself or e^6x by itself we could do the integral easily.  Although, we don't have them by themselves, they are in place of multiplied together.

There is no substitution that we can use on this integral that will allow us to do the integral.  So, at this point we don't have the knowledge to do this integral.

To do this integral we will require to make use of integration by parts so let's derive the integration by parts formula. We'll begin with the product rule.

(f g)′ = f'g + f g′

Here, integrate both sides of this.

∫ (f g)′ dx = ∫ f ′ g + f g′ dx

The left side is very easy to integrate and we'll divide the right side of the integral.

Fg = ∫ f' g dx + ∫ fg'dx

Note: Technically we should comprise had a constant of integration show up on the left side later than doing the integration. We can drop it at this point as other constants of integration will be showing up down the road and they would just end up absorbing this one.

At last, rewrite the formula as follows and we arrive at the integration by parts formula.

∫ f g′ dx = fg - ∫ f ′ g dx

Though, this is not the easy formula to use.  Thus, let's do a couple of substitutions.

u = f (x)

v = g (x)

du = f ′ (x) dx

dv = g ′ (x) dx

Both of these formulas are just the standard Calc I substitutions which hopefully you are used to by now. Don't get excited by the fact that we are by using 2 substitutions here. They will work similar way.

By using these substitutions provides us the formula that most people think of as the integration by parts formula.

∫ u dv = uv - ∫ v du

To employ this formula we will require identifying u and dv, calculating du and v and then using the formula. Note also that computing v is very easy.  All we require to do is integrate dv.

v = ∫ dv

So, let's take a look at the integral above that we specified we wanted to do.