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**preview**shows page 1. to view the full**5 pages of the document.**MA 141 Chapter 4

Section 4.5: Integration by Parts

Just like substitution is linked to the Chain Rule, we have a method for dealing with integrals

related to the Product Rule.

Recall: (fg)′(x) = f(x)g′(x) + f′(x)g(x).

We can rearrange this equation as follows:

f(x)g′(x) = (fg)′(x)−f′(x)g(x)

Integrating both sides, we get

Zf(x)g′(x)dx =Z(fg)′(x)dx −Zf′(x)g(x)dx

=f(x)g(x)−Zf′(x)g(x)dx

Using substitution, we can make this look simpler. Let u=f(x)and v=g(x).

Zu dv =uv −Zv du

Given an integral where we want to use integration by parts, we need to assign something to

uand dv. In general, choose the easier function to integrate for dv and the easier function

to derive for u.

The mnemonic ETAIL gives an increasing diﬃculty in integration:

1) Exponential

2) Trigonometric

3) Algebraic

4) Inverse trigonometric

5) Logarithmic

Try to pick an easier type to integrate, like exponential or trigonometric, if possible.

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Ex: Find Zxsec2xdx.

Notice that we cannot choose u=xor u= sec2xand use substitution.

We’ll use integration by parts. We have no exponential piece, so we will choose

dv = sec2xdx. Then u=x.

Diﬀerentiate Integrate

u=x dv = sec2xdx

du =dx v = tan x

Do not include +Cin the integration of dv - save this for the end when all integration is

done.

Zxsec2xdx =Zudv

=uv −Zvdu

=xtan x−Ztan xdx

=xtan x+ ln |cos x|+C

Ex: Find Zx3ln(x)dx.

We don’t have an exponential or a trig function, so we’ll choose dv =x3dx and u= ln(x).

u= ln(x)dv =x3dx

du =1

xdx v =x4

4

Zx3ln(x)dx =uv −Zv du

=x4

4ln(x)−Zx4

4·1

xdx

=x4

4ln(x)−1

4Zx3dx

=x4

4ln(x)−x4

16 +C

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