1. The following integration can be solved by using the Substitution f 14 dx = ♦ (Choose the correct letter). A. 2ln+c B. 2+2 ln |x²|+c C. 2 In 1 + x²|+c D. 2 tan-¹+c E. None of these are correct 2. The following integration can be solved by using the fdx = ◆ (Choose the correct letter). A. 2 tan¹x+c B. 2 In 1+x+c C. 2 ln |1+x³ + c D. 2 tan-¹³ + c E. None of these are correct 3. The following integration can be solved by using the ◆ , and du= f 2x sinx dx = ♦ (Choose the correct letter), to get ♦ technique, where we have = → technique, where we have u technique, where we have = ◆ and du and du= du =

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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question 4
edu.bh/moodle/mod/quiz/attempt.php?attempt=41667&amid=177639
1. The following integration can be solved by using the Substitution
f 14 dx =
(Choose the correct letter).
1+²
A. 2 ln ||+c
B. 2+2 ln |x²|+c
C. 2 ln |1 + x²|+c
D. 2 tan-¹x+c
E. None of these are correct
2. The following integration can be solved by using the
fb dx =
(Choose the correct letter).
1+26
A. 2 tan¹x+c
B. 2 ln |1+x+c
C. 2 In 1 + x³ + c
D.
2 tan-¹³ + c
E. None of these are correct
3. The following integration can be solved by using the
and du
◆
f 2x sin x dx =
(Choose the correct letter),
O Hi
re to search
to get
a
♦ technique, where we have u=
→
technique, where we have u=
technique, where we have u =
and du=
and du=
du =
DE
56
Transcribed Image Text:edu.bh/moodle/mod/quiz/attempt.php?attempt=41667&amid=177639 1. The following integration can be solved by using the Substitution f 14 dx = (Choose the correct letter). 1+² A. 2 ln ||+c B. 2+2 ln |x²|+c C. 2 ln |1 + x²|+c D. 2 tan-¹x+c E. None of these are correct 2. The following integration can be solved by using the fb dx = (Choose the correct letter). 1+26 A. 2 tan¹x+c B. 2 ln |1+x+c C. 2 In 1 + x³ + c D. 2 tan-¹³ + c E. None of these are correct 3. The following integration can be solved by using the and du ◆ f 2x sin x dx = (Choose the correct letter), O Hi re to search to get a ♦ technique, where we have u= → technique, where we have u= technique, where we have u = and du= and du= du = DE 56
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