Calculus II Exam 02 Review September 30, 2021
Directions: Show all work to receive full credit.
Name:
1. Solve the following integrals.
(a) ∫
x (2x+1)3
dx
1 4 ( 1 2(2x+1)2
− 1 2x+1
)
(b) ∫ x cos2(x) dx
x 2 (x +
sin(2x) 2
)− 1 2 (x
2
2 − cos(2x)
4 )
(c) ∫
ln( √ x)
x dx
1 4 (ln(x))2
(d) ∫ x sin(x) cos(x) dx
1 2 (−x
2 cos(2x) + 1
4 sin(2x))
Calculus II Exam 02 Review Page 2 of 5
(e) ∫
1 (x2−1)2 dx
− ln | x√ x2−1
+ 1√ x2−1 |
(f) ∫ x √ x−3 dx
2 5 (x−3)5/2 + 2(x−3)3/2 + C
(g) ∫ x2 ln(x) dx
x3
3 ln(x)− 1
9 x3 + C
(h) ∫ √
1+x2
x dx
ln | √ 1+x2
x − 1
x |+ √ x2 + 1 + C
Cont.
Calculus II Exam 02 Review Page 3 of 5
(i) ∫
ex
1+ex dx
ln(ex+1) Think about why there is no absolute value bars
(j) ∫ sin2(x) cos2(x) dx
1 8 (x− sin(4x)
4 ) + C
(k) ∫
x2+1√ x2+2x+2
dx (Note: This took me almost an entire page. Not Mandatory.)
( √
(x+1)2+1)(x+1)+3 ln | √
(x2+1)2+1+(x+1)| 2
−2 √ (x + 1)2 + 1
(l) ∫
x−4 x2−5x+6 dx
− ln |x−3|+ 2 ln |x−2|+ C
Cont.
Calculus II Exam 02 Review Page 4 of 5
(m) ∫ tan2(x) sec4(x) dx
tan5(x) 5
+ tan3(x)
3 + C
(n) ∫
ln( √ x)√ x
dx
2( √ x ln( √ x)−
√ x) + C
(o) ∫ arctan(x) dx
x arctan(x)− 1 2 ln(1+x2)+C Why not absolute value?
(p) ∫ sin3(x) cos2(x) dx
−(cos 3(x) 3 − cos
5(x) 5
) + C
Cont.
Calculus II Exam 02 Review Page 5 of 5
2. Use Simpson’s rule to estimate the integral. ∫ 2 0
x2
x2+1 dx where n = 8
Approximately: .8928598757
3. Use the Trapazoid rule to estimate the integral. ∫ 3 1
sin(x) x
dx where n = 4
Approximately: .90164486
4. Determine the number of subintervals are needed to estimate the integral ∫ 2 0 xex dx accurate to within
0.001 units using Simpson’s rule.
n ≥ 10
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