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### Integration By Parts

• Know by heart:
• Choosing uand dv:
• dvcan be integrated
• is at most “just as difficult” as original integral
• If the first integration is not successful, apply integration by parts again

### Integration By Parts

2xcosxdx
• Let u = 2x and dv = cosx
• du = 2dx
• dv = cosxdx
• v = sinx + C
Apply Integration by Parts
2xcosxdx = 2x(sinx) − ∫ sinx(2dx) + C = 2xsinx − 2∫ sinxdx + C = 2xsinx + 2cosx + C
xsin3xdx
• Let u = x and dv = sin3xdx
• u = x
• du = dx
• dv = sin3xdx
• v = [( − cos3x)/3] + C
• Apply Integration by Parts
• xsin3xdx = [( − cos3x)/3]( x ) − ∫ [( − cos3x)/3]dx + C
• = [( − xcos3x)/3] + [1/3]∫ cos3xdx + C
• = [( − xcos3x)/3] + [1/3]( [sin3x/3] ) + C
• = [( − xcos3x)/3] + [sin3x/9] + C
= [( − 3xcos3x + sin3x)/9] + C
lnx2dx
• Use logarithm properties to simplify
• lnx2dx = ∫ 2lnxdx
• = 2∫ lnxdx
• Let u = lnx and dv = dx
• du = [dx/x]
• dv = dx
• v = x + C
• Apply Integration by Parts
• lnx2dx = 2∫ lnxdx
= 2( xlnx − ∫ dx + C ) = 2( xlnx − x + C ) = 2xlnx − 2x + C
ln(x + 1)dx
• Let u = x + 1 and dv = dx
• u = ln(x + 1)
• du = [1/(x + 1)]dx
• dv = dx
• v = x + C
• Apply Integration by Parts
• ln(x + 1)dx = x( ln(x + 1) ) − ∫ x( [dx/(3x + 1)] ) + C
• = x( ln(x + 1) ) − ∫ [x/(x + 1)] dx + C
• Note the alternative form of [x/(x + 1)]
• [x/(x + 1)] = 1 − [1/(x + 1)]
• x( ln(x + 1) ) − ∫ [x/(x + 1)] dx = x( ln(x + 1) ) − ∫ 1 − [1/(x + 1)] dx + C
• = xln(x + 1) − ∫ dx + ∫ [1/(x + 1)] dx + C
= xln(x + 1) − x + ln(x + 1) + C
3x2 exdx
• Let u = x2 and dv = exdx
• du = 2x dx
• dv = exdx
• v = ex + C
• Apply Integration by Parts
• 3x2 exdx = 3∫ x2 exdx
• = 3( exx2 − ∫ ex( 2x dx ) )
• Apply Integration by Parts again on ∫ ex( 2x dx )
• ex( 2x dx ) = ex(2x) + ∫ ex(2dx)
• = 2xex + 2ex + C
Apply substitution
3( exx2 − ∫ ex( 2x dx ) ) = 3( exx2 − 2xex + 2ex + C )
0π (x − 1)sinxdx
• Let u = x − 1 and dv = sinxdx
• du = dx
• dv = sinxdx
• v = − cosx
• Apply Integration by Parts
• 0π (x − 1)sinxdx = − cosx(x − 1) − ∫0π− cosx(dx)
• = (cosx − xcosx + sinx)0π
• = cosπ− πcos π+ sinπ− (cos0 − (0)cos0 + sin0)
• = − 1 + π+ 0 − (1)
= π− 2
010 x exdx
• Let u = x and dv = exdx
• du = dx
• dv = exdx
• v = ex
• Apply Integration by Parts
• 010 x exdx = [ ex(x) ]010 + ∫010 exdx
= e10(10) + e10 − (e0(0) + e0) = e10(10) + e10 − 1 = 11e10 − 1
0e [x/(x + 1)]dx
• Let u = x and dv = [dx/(x + 1)]
• du = dx
• dv = [1/(x + 1)]dx
• v = ln(x + 1)
• Apply Integration by Parts
• 0e [x/(x + 1)]dx = xln(x + 1) − ∫0e ln(x + 1) dx
• From problem 4, we have solved ∫ ln(x + 1) dx
• ln(x + 1) dx = xln(x + 1) − x + ln(x + 1) + C
• xln(x + 1) − ∫0e ln(x + 1) dx = xln(x + 1) − (xln(x + 1) − x + ln(x + 1))
• = [ xln(x + 1) − xln(x + 1) + x − ln(x + 1) ]0e
• = eln(e + 1) − eln(e + 1) + e − ln(e + 1) − ((0)ln(0 + 1) − (0)ln(0 + 1) + 0 − ln(0 + 1)
• = e − ln(e + 1) + ln1
= e − ln(e + 1)
23 ln(3x − 5)3dx
• Use logarithm properties to change form
• 23 ln(3x − 5)3dx = ∫23 3ln(3x − 5)dx
• = ∫02 3n(3x − 5)3dx
• Let u = ln(3x − 5) and dv = 3dx
• du = [3/(3x − 5)]dx
• dv = 3dx
• v = 3x
• Apply Integration by Parts
• 23 ln(3x − 5)dx = xln(3x − 5) − ∫23 [3x/(3x − 5)]dx
• Find the alternative form of [3x/(3x − 5)]
• [3x/(3x − 5)] = [(3x − 5)/(3x − 5)] + [5/(3x − 5)] = 1 + [5/(3x − 5)]
• xln(3x − 5) − ∫02 [3x/(3x − 5)]dx = xln(3x − 5) − ∫23 1 + [5/(3x − 5)] dx
= 3ln4 − 3 − [5/3]ln4 − (2ln1 − 2 − [5/3]ln1) = 3ln4 − [5/3]ln4 − 2ln1 + [5/3]ln1 − 1 = [4/3]ln4 − 1
Prove ∫ sinxcosx = [(sin2x)/2] + C
• Let u = sinx and dv = cosx
• du = cosx
• dv = cosx
• v = sinx
• Apply Integration by Parts
• sinxcosx = sinx(sinx) − ∫ sinxcosx + C
• = sin2x − ∫ sinxcosx + C
• Isolate sin2x
• 2∫ sinxcosx = sin2x + C
sinxcosx = [(sin2x)/2] + C

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Integration By Parts

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Integration by Parts 0:12
• Formula
• How It's Derived From Product Rule
• Integration by Parts: Rules to Follow 3:24
• Recap of the Rules
• Example 1 4:29
• Example 2 7:04
• Example 3 8:40
• Example 4 13:48
• Example 5 15:36