Exercise 7.5 answers: Integrals

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1. The denominator already has two different linear factors, x + 1 and x + 2, so write x/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
2. Multiply both sides by (x + 1)(x + 2): x = A(x + 2) + B(x + 1).
3. Put x = −1: −1 = A(1), so A = −1.
4. Put x = −2: −2 = B(−1), so B = 2.
5. So x/((x + 1)(x + 2)) = −1/(x + 1) + 2/(x + 2).
6. Integrate each piece: −log|x + 1| + 2 log|x + 2|.

Answer2 log|x + 2| − log|x + 1| + C

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1. Factor the denominator: x² − 9 = (x − 3)(x + 3).
2. Write 1/((x − 3)(x + 3)) = A/(x − 3) + B/(x + 3).
3. Multiply out: 1 = A(x + 3) + B(x − 3).
4. Put x = 3: 1 = 6A, so A = 1/6.
5. Put x = −3: 1 = −6B, so B = −1/6.
6. Integrate each piece: (1/6) log|x − 3| − (1/6) log|x + 3|.

Answer(1/6) log|(x − 3)/(x + 3)| + C

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1. Three different linear factors, so write (3x − 1)/((x − 1)(x − 2)(x − 3)) = A/(x − 1) + B/(x − 2) + C/(x − 3).
2. Multiply out: 3x − 1 = A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2).
3. Put x = 1: 2 = A(−1)(−2) = 2A, so A = 1.
4. Put x = 2: 5 = B(1)(−1) = −B, so B = −5.
5. Put x = 3: 8 = C(2)(1) = 2C, so C = 4.
6. Integrate each piece: log|x − 1| − 5 log|x − 2| + 4 log|x − 3|.

Answerlog|x − 1| − 5 log|x − 2| + 4 log|x − 3| + C

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1. Write x/((x − 1)(x − 2)(x − 3)) = A/(x − 1) + B/(x − 2) + C/(x − 3).
2. Multiply out: x = A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2).
3. Put x = 1: 1 = A(−1)(−2) = 2A, so A = 1/2.
4. Put x = 2: 2 = B(1)(−1) = −B, so B = −2.
5. Put x = 3: 3 = C(2)(1) = 2C, so C = 3/2.
6. Integrate each piece: (1/2) log|x − 1| − 2 log|x − 2| + (3/2) log|x − 3|.

Answer(1/2) log|x − 1| − 2 log|x − 2| + (3/2) log|x − 3| + C

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1. Factor the denominator: x² + 3x + 2 = (x + 1)(x + 2).
2. Write 2x/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
3. Multiply out: 2x = A(x + 2) + B(x + 1).
4. Put x = −1: −2 = A(1), so A = −2.
5. Put x = −2: −4 = B(−1), so B = 4.
6. Integrate each piece: −2 log|x + 1| + 4 log|x + 2|.

Answer4 log|x + 2| − 2 log|x + 1| + C

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1. The top and bottom both have degree 2 (bottom: x(1 − 2x) = x − 2x²), so this fraction is improper — divide first.
2. Dividing 1 − x² by x − 2x² gives a quotient of 1/2, with remainder 1 − x/2. So the fraction equals 1/2 + (1 − x/2)/(x(1 − 2x)).
3. Split the remaining proper fraction: (1 − x/2)/(x(1 − 2x)) = A/x + B/(1 − 2x).
4. Multiply out: 1 − x/2 = A(1 − 2x) + Bx.
5. Put x = 0: 1 = A, so A = 1.
6. Compare the x-coefficient: −1/2 = −2A + B = −2 + B, so B = 3/2.
7. Integrate: 1/2 dx gives x/2; 1/x gives log|x|; (3/2)/(1 − 2x) gives −(3/4) log|1 − 2x|.
8. Add all three pieces.

Answerx/2 + log|x| − (3/4) log|1 − 2x| + C

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1. The denominator has one linear factor (x − 1) and one quadratic factor that does not break further (x² + 1), so write x/((x² + 1)(x − 1)) = A/(x − 1) + (Bx + C)/(x² + 1).
2. Multiply out: x = A(x² + 1) + (Bx + C)(x − 1).
3. Put x = 1: 1 = 2A, so A = 1/2.
4. Compare the x² coefficient: 0 = A + B, so B = −1/2.
5. Compare the constant term: 0 = A − C, so C = 1/2.
6. So the fraction is (1/2)/(x − 1) + (−x/2 + 1/2)/(x² + 1).
7. Integrate: (1/2) log|x − 1|, then split −x/(2(x² + 1)) + 1/(2(x² + 1)) to get −(1/4) log(x² + 1) + (1/2) tan⁻¹x.
8. Add the pieces together.

Answer(1/2) log|x − 1| − (1/4) log(x² + 1) + (1/2) tan⁻¹x + C

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1. The factor (x − 1) is repeated, so it needs two terms: x/((x − 1)²(x + 2)) = A/(x − 1) + B/(x − 1)² + C/(x + 2).
2. Multiply out: x = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
3. Put x = 1: 1 = 3B, so B = 1/3.
4. Put x = −2: −2 = 9C, so C = −2/9.
5. Compare the x² coefficient: 0 = A + C, so A = 2/9.
6. So the fraction is (2/9)/(x − 1) + (1/3)/(x − 1)² + (−2/9)/(x + 2).
7. Integrate: (2/9) log|x − 1| − 1/(3(x − 1)) − (2/9) log|x + 2|.

Answer(2/9) log|x − 1| − (2/9) log|x + 2| − 1/(3(x − 1)) + C

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1. Factor the denominator: x³ − x² − x + 1 = x²(x − 1) − (x − 1) = (x − 1)(x² − 1) = (x − 1)²(x + 1).
2. The factor (x − 1) is repeated, so write (3x + 5)/((x − 1)²(x + 1)) = A/(x − 1) + B/(x − 1)² + C/(x + 1).
3. Multiply out: 3x + 5 = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)².
4. Put x = 1: 8 = 2B, so B = 4.
5. Put x = −1: 2 = 4C, so C = 1/2.
6. Compare the x² coefficient: 0 = A + C, so A = −1/2.
7. So the fraction is (−1/2)/(x − 1) + 4/(x − 1)² + (1/2)/(x + 1).
8. Integrate: −(1/2) log|x − 1| − 4/(x − 1) + (1/2) log|x + 1|.

Answer(1/2) log|x + 1| − (1/2) log|x − 1| − 4/(x − 1) + C

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1. Factor x² − 1 = (x − 1)(x + 1), giving three distinct linear factors in all.
2. Write (2x − 3)/((x − 1)(x + 1)(2x + 3)) = A/(x − 1) + B/(x + 1) + C/(2x + 3).
3. Multiply out: 2x − 3 = A(x + 1)(2x + 3) + B(x − 1)(2x + 3) + C(x − 1)(x + 1).
4. Put x = 1: −1 = 10A, so A = −1/10.
5. Put x = −1: −5 = −2B, so B = 5/2.
6. Put x = −3/2: −6 = (5/4)C, so C = −24/5.
7. So the fraction is (−1/10)/(x − 1) + (5/2)/(x + 1) + (−24/5)/(2x + 3).
8. Integrate: −(1/10) log|x − 1| + (5/2) log|x + 1| − (12/5) log|2x + 3| (dividing by the extra factor of 2 inside 2x + 3).

Answer−(1/10) log|x − 1| + (5/2) log|x + 1| − (12/5) log|2x + 3| + C

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1. Factor x² − 4 = (x − 2)(x + 2), giving three distinct linear factors in all.
2. Write 5x/((x + 1)(x − 2)(x + 2)) = A/(x + 1) + B/(x − 2) + C/(x + 2).
3. Multiply out: 5x = A(x − 2)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x − 2).
4. Put x = −1: −5 = −3A, so A = 5/3.
5. Put x = 2: 10 = 12B, so B = 5/6.
6. Put x = −2: −10 = 4C, so C = −5/2.
7. So the fraction is (5/3)/(x + 1) + (5/6)/(x − 2) + (−5/2)/(x + 2).
8. Integrate each piece and add.

Answer(5/3) log|x + 1| + (5/6) log|x − 2| − (5/2) log|x + 2| + C

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1. The top has degree 3 and the bottom has degree 2, so this fraction is improper — divide first.
2. Dividing x³ + x + 1 by x² − 1 gives quotient x, remainder 2x + 1. So the fraction equals x + (2x + 1)/(x² − 1).
3. Factor x² − 1 = (x − 1)(x + 1) and split: (2x + 1)/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1).
4. Multiply out: 2x + 1 = A(x + 1) + B(x − 1).
5. Put x = 1: 3 = 2A, so A = 3/2.
6. Put x = −1: −1 = −2B, so B = 1/2.
7. Integrate: ∫x dx gives x²/2; the two partial fractions give (3/2) log|x − 1| + (1/2) log|x + 1|.
8. Add all the pieces.

Answerx²/2 + (3/2) log|x − 1| + (1/2) log|x + 1| + C

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1. Write 1 − x = −(x − 1), so the fraction is 2/((1 − x)(1 + x²)) = −2/((x − 1)(x² + 1)).
2. The denominator has one linear factor and one quadratic factor that will not break, so write it as A/(x − 1) + (Bx + C)/(x² + 1).
3. Multiply out: −2 = A(x² + 1) + (Bx + C)(x − 1).
4. Put x = 1: −2 = 2A, so A = −1.
5. Compare the x² coefficient: 0 = A + B, so B = 1.
6. Compare the constant term: −2 = A − C, so C = 1.
7. So the fraction is −1/(x − 1) + (x + 1)/(x² + 1).
8. Integrate: −log|x − 1|, then split x/(x² + 1) + 1/(x² + 1) to get (1/2) log(x² + 1) + tan⁻¹x.
9. Add the pieces together.

Answer−log|x − 1| + (1/2) log(x² + 1) + tan⁻¹x + C

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1. The whole denominator is one repeated factor, (x + 2)², so write (3x − 1)/(x + 2)² = A/(x + 2) + B/(x + 2)².
2. Multiply out: 3x − 1 = A(x + 2) + B.
3. Put x = −2: −7 = B, so B = −7.
4. Compare the x-coefficient: 3 = A.
5. So the fraction is 3/(x + 2) + (−7)/(x + 2)².
6. Integrate: 3 log|x + 2|, and ∫−7/(x + 2)² dx = 7/(x + 2).
7. Add the two pieces.

Answer3 log|x + 2| + 7/(x + 2) + C

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1. Factor the denominator: x⁴ − 1 = (x² − 1)(x² + 1) = (x − 1)(x + 1)(x² + 1).
2. Write 1/((x − 1)(x + 1)(x² + 1)) = A/(x − 1) + B/(x + 1) + (Cx + D)/(x² + 1).
3. Multiply out and put x = 1: 1 = 4A, so A = 1/4.
4. Put x = −1: 1 = −4B, so B = −1/4.
5. Comparing the x³ and x² coefficients gives C = 0 and D = −1/2.
6. So the fraction is (1/4)/(x − 1) + (−1/4)/(x + 1) + (−1/2)/(x² + 1).
7. Integrate each piece: (1/4) log|x − 1| − (1/4) log|x + 1| − (1/2) tan⁻¹x.

Answer(1/4) log|x − 1| − (1/4) log|x + 1| − (1/2) tan⁻¹x + C

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1. Follow the hint: multiply the top and bottom by xⁿ⁻¹. This gives xⁿ⁻¹/(xⁿ(xⁿ + 1)) dx.
2. Put t = xⁿ. Differentiating, dt = n·xⁿ⁻¹ dx, so xⁿ⁻¹ dx = dt/n.
3. The integral becomes (1/n) ∫ dt/(t(t + 1)).
4. Split 1/(t(t + 1)) = 1/t − 1/(t + 1) (put t = 0 to get the first constant as 1, and t = −1 to get the second as −1).
5. Integrate: (1/n)[log|t| − log|t + 1|] = (1/n) log|t/(t + 1)|.
6. Put back t = xⁿ.

Answer(1/n) log|xⁿ/(xⁿ + 1)| + C

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1. The bottom has sin x in it, and the top is cos x — exactly the rate of change of sin x. Put t = sin x, so dt = cos x dx.
2. The integral becomes ∫ dt/[(1−t)(2−t)].
3. Split into simple fractions: 1/[(1−t)(2−t)] = 1/(1−t) − 1/(2−t).
4. Integrate each piece: ∫1/(1−t) dt = −log|1−t|, and ∫1/(2−t) dt = −log|2−t|.
5. So in t the integral is −log|1−t| − (−log|2−t|) + C = log|2−t| − log|1−t| + C.
6. Put back t = sin x: log|2 − sin x| − log|1 − sin x| + C.

Answerlog|2 − sin x| − log|1 − sin x| + C

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1. Top and bottom both have degree 4 in x, so this fraction is improper — divide first.
2. Write y = x². The fraction is (y+1)(y+2)/[(y+3)(y+4)] = (y²+3y+2)/(y²+7y+12).
3. Dividing: (y²+3y+2)/(y²+7y+12) = 1 − (4y+10)/[(y+3)(y+4)].
4. Split the proper part: (4y+10)/[(y+3)(y+4)] = −2/(y+3) + 6/(y+4).
5. So the integrand is 1 + 2/(x²+3) − 6/(x²+4).
6. Integrate term by term: ∫1 dx = x, ∫2/(x²+3) dx = (2/√3) tan⁻¹(x/√3), ∫6/(x²+4) dx = 3 tan⁻¹(x/2).
7. Add the pieces: x + (2/√3) tan⁻¹(x/√3) − 3 tan⁻¹(x/2) + C.

Answerx + (2/√3) tan⁻¹(x/√3) − 3 tan⁻¹(x/2) + C

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1. The top, 2x, is exactly the rate of change of x², which sits inside both factors below. Put t = x², so dt = 2x dx.
2. The integral becomes ∫ dt/[(t+1)(t+3)].
3. Split: 1/[(t+1)(t+3)] = (1/2)[1/(t+1) − 1/(t+3)].
4. Integrate: (1/2)[log|t+1| − log|t+3|] + C.
5. Put back t = x²: (1/2) log(x²+1) − (1/2) log(x²+3) + C.

Answer(1/2) log(x²+1) − (1/2) log(x²+3) + C

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1. Multiply the top and bottom by x³: 1/[x(x⁴−1)] = x³/[x⁴(x⁴−1)].
2. Now x⁴ appears everywhere, and its rate of change 4x³ is (up to a constant) the new top. Put t = x⁴, so dt = 4x³ dx.
3. The integral becomes (1/4)∫ dt/[t(t−1)].
4. Split: 1/[t(t−1)] = 1/(t−1) − 1/t.
5. Integrate: (1/4)[log|t−1| − log|t|] + C.
6. Put back t = x⁴: (1/4) log|x⁴−1| − (1/4) log(x⁴) + C, and log(x⁴) = 4 log|x|, giving (1/4) log|x⁴−1| − log|x| + C.

Answer(1/4) log|x⁴ − 1| − log|x| + C

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1. Put t = eˣ, so dt = eˣ dx, that is dx = dt/t.
2. The integral becomes ∫ dt/[t(t−1)].
3. Split: 1/[t(t−1)] = 1/(t−1) − 1/t.
4. Integrate: log|t−1| − log|t| + C.
5. Put back t = eˣ: log|eˣ−1| − log(eˣ) + C, and log(eˣ) = x, giving log|eˣ−1| − x + C.

Answerlog|eˣ − 1| − x + C

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1. Split x/[(x−1)(x−2)] into simple fractions A/(x−1) + B/(x−2).
2. Cover-up at x=1: A = 1/(1−2) = −1. At x=2: B = 2/(2−1) = 2.
3. So the integrand is −1/(x−1) + 2/(x−2).
4. Integrate: −log|x−1| + 2 log|x−2| + C = log[(x−2)²/|x−1|] + C.
5. This matches option (B).

Answer(B) log|(x−2)²/(x−1)| + C

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1. Split 1/[x(x²+1)] as A/x + (Bx+C)/(x²+1).
2. Clearing denominators: 1 = A(x²+1) + (Bx+C)x. Matching constants gives A=1; matching x² gives A+B=0 so B=−1; matching x gives C=0.
3. So the integrand is 1/x − x/(x²+1).
4. Integrate: log|x| − (1/2) log(x²+1) + C.
5. This matches option (A).

Answer(A) log|x| − (1/2) log(x²+1) + C

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