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Calculus II Exam 02 Review September 30, 2021

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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

The End.

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