3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Throughout, C denotes an arbitrary constant of integration. On R, ∫ex(x2+1)2x3−x+2dx is ______. Exactly one option is correct.
Aexx2+1x−1+C
Bexx2+1x+C
Cexx2+1x+1+C
Dex(x2+1)2x2+x+C
Show answer and solution
Answer: C
Look for f with f+f′=(x2+1)2x3−x+2, because dxd(exf)=ex(f+f′).
The denominator suggests f=x2+1x+1: then f′=(x2+1)2(x2+1)−2x(x+1)=(x2+1)21−2x−x2 and
f+f′=(x2+1)2(x+1)(x2+1)+1−2x−x2=(x2+1)2x3−x+2.
Hence the integral is exx2+1x+1+C on R (one interval, so one constant).
Check by differentiation: dxd[exx2+1x+1]=ex(f+f′), which is the integrand. The other options are not antiderivatives: at x=0 the integrand equals 2, while the derivatives of (A), (B), (D) at x=0 equal 0, 1, 1 respectively.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
Throughout, C denotes an arbitrary constant of integration. For x>0, ∫x2(x4+1)3/4dx is ______. Exactly one option is correct.
A−(1+x41)1/4+C
B(1+x41)1/4+C
C−x2(1+x4)1/4+C
D−41(1+x41)1/4+C
Show answer and solution
Answer: A
For x>0, (x4+1)3/4=x3(1+x−4)3/4, so the integrand is (1+x−4)3/4x−5.
Put u=1+x−4, du=−4x−5dx: the integral becomes −41∫u−3/4du=−u1/4+C=−(1+x41)1/4+C.
Check: dxd[−(1+x−4)1/4]=−41(1+x−4)−3/4(−4x−5)=(1+x−4)3/4x−5=x2(x4+1)3/41.
Option (B) has the wrong sign, (D) the wrong factor 41, and (C) equals −x(1+x−4)1/4, whose derivative is different.
Difficulty: Hard
Question 3Level 2 · ExamOne or more correct
Let f(x)=xx2−11 for x>1. Which of the following functions is (are) antiderivatives of f on the interval (1,∞)? (All inverse trigonometric functions take their principal values.) One or more options may be correct.
Asin−1x1
Bsec−1x
Ctan−1x2−1
D−2tan−1(x−x2−1)
Show answer and solution
Answer: B, C, D
Differentiate each on (1,∞).
(A) dxdsin−1x1=1−1/x21⋅(−x21)=−xx2−11: the negative of f. False.
(B) dxdsec−1x=xx2−11 for x>1. True.
(C) dxdtan−1x2−1=x21⋅x2−1x=xx2−11. True.
(D) Put x=secθ, θ∈(0,2π). Then x−x2−1=cosθ1−sinθ=tan(4π−2θ) with 4π−2θ∈(0,4π), so −2tan−1(x−x2−1)=θ−2π=sec−1x−2π, which differs from (B) by a constant. True.
Answer: B, C, D.
Difficulty: Hard
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