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Final

# MA 141 Lecture 25: Integration by PartsExam

Department
Mathematics
Course Code
MA 141
Professor
Kevin Flores
Study Guide
Final

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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 xZtan 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