Exercise 7.4 answers: Integrals

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1. Notice the denominator has x⁶, and the numerator has x², which is (1/3) of the derivative of x³.
2. Substitute t = x³, so dt = 3x² dx.
3. The integral becomes ∫ dt/(t² + 1), a standard form.
4. ∫ dt/(t² + 1) = tan⁻¹t + C.
5. Put back t = x³: the integral is tan⁻¹(x³) + C.

Answertan⁻¹(x³) + C

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1. Write 4x² as (2x)². Let t = 2x, so dt = 2 dx, that is dx = dt/2.
2. The integral becomes (1/2) ∫ dt/√(t² + 1).
3. Use the standard form ∫dt/√(t² + a²) = log|t + √(t² + a²)| + C, with a = 1.
4. This gives (1/2) log|t + √(t² + 1)| + C.
5. Put back t = 2x: (1/2) log|2x + √(4x² + 1)| + C.

Answer(1/2) log|2x + √(4x² + 1)| + C

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1. Let t = x − 2 (so (2 − x)² = t² and dx = dt).
2. The integral becomes ∫ dt/√(t² + 1).
3. Use ∫dx/√(x²+a²) = log|x + √(x²+a²)| + C, with a = 1.
4. This gives log|t + √(t² + 1)| + C.
5. Put back t = x − 2: log|(x − 2) + √((2 − x)² + 1)| + C.

Answerlog|(x − 2) + √((2 − x)² + 1)| + C

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1. Write 9 − 25x² as 25((3/5)² − x²), so the square root is 5√((3/5)² − x²).
2. The integral becomes (1/5) ∫ dx/√((3/5)² − x²).
3. Use the standard form ∫dx/√(a² − x²) = sin⁻¹(x/a) + C, with a = 3/5.
4. This gives (1/5) sin⁻¹(x ÷ (3/5)) + C.
5. Simplify: (1/5) sin⁻¹(5x/3) + C.

Answer(1/5) sin⁻¹(5x/3) + C

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1. The power x⁴ suggests t = x², so dt = 2x dx, i.e. x dx = dt/2.
2. The integral becomes (3/2) ∫ dt/(1 + 2t²).
3. Write 1 + 2t² = 2(t² + 1/2), so this is (3/4) ∫ dt/(t² + 1/2).
4. Use ∫dt/(t²+a²) = (1/a) tan⁻¹(t/a) + C, with a = 1/√2.
5. This gives (3/4)·√2·tan⁻¹(√2 t) + C = (3√2/4) tan⁻¹(√2 x²) + C.

Answer(3√2/4) tan⁻¹(√2 x²) + C

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1. Notice x⁶ = (x³)², and the numerator x² is (1/3) the derivative of x³.
2. Substitute t = x³, so dt = 3x² dx.
3. The integral becomes (1/3) ∫ dt/(1 − t²).
4. Use ∫dt/(1−t²) = (1/2) log|(1+t)/(1−t)| + C.
5. This gives (1/6) log|(1 + x³)/(1 − x³)| + C.

Answer(1/6) log|(1 + x³)/(1 − x³)| + C

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1. Split the fraction: (x − 1)/√(x²−1) = x/√(x²−1) − 1/√(x²−1).
2. For the first part, ∫x/√(x²−1) dx = √(x² − 1), since the numerator is half the derivative of x²−1.
3. For the second part, use ∫dx/√(x²−a²) = log|x + √(x²−a²)| + C, with a = 1.
4. This gives log|x + √(x² − 1)| + C.
5. Together: √(x² − 1) − log|x + √(x² − 1)| + C.

Answer√(x² − 1) − log|x + √(x² − 1)| + C

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1. Notice x⁶ = (x³)² and a⁶ = (a³)², and the numerator x² is (1/3) the derivative of x³.
2. Substitute t = x³, so dt = 3x² dx.
3. The integral becomes (1/3) ∫ dt/√(t² + (a³)²).
4. Use ∫dt/√(t²+k²) = log|t + √(t²+k²)| + C, with k = a³.
5. This gives (1/3) log|x³ + √(x⁶ + a⁶)| + C.

Answer(1/3) log|x³ + √(x⁶ + a⁶)| + C

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1. Let t = tan x, so dt = sec²x dx.
2. The integral becomes ∫ dt/√(t² + 4).
3. Use ∫dt/√(t²+a²) = log|t + √(t²+a²)| + C, with a = 2.
4. This gives log|t + √(t² + 4)| + C.
5. Put back t = tan x: log|tan x + √(tan²x + 4)| + C.

Answerlog|tan x + √(tan²x + 4)| + C

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1. Complete the square: x² + 2x + 2 = (x + 1)² + 1.
2. Let t = x + 1, so the integral is ∫ dt/√(t² + 1).
3. Use ∫dt/√(t²+a²) = log|t + √(t²+a²)| + C, with a = 1.
4. This gives log|t + √(t² + 1)| + C.
5. Put back t = x + 1: log|(x + 1) + √(x² + 2x + 2)| + C.

Answerlog|(x + 1) + √(x² + 2x + 2)| + C

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1. Complete the square: 9x² + 6x + 5 = 9(x + 1/3)² + 4.
2. Let t = x + 1/3, so the integral is ∫ dt/(9t² + 4) = (1/9) ∫ dt/(t² + 4/9).
3. Use ∫dt/(t²+a²) = (1/a) tan⁻¹(t/a) + C, with a = 2/3.
4. This gives (1/9)·(3/2)·tan⁻¹(3t/2) + C = (1/6) tan⁻¹(3t/2) + C.
5. Put back t = x + 1/3: (1/6) tan⁻¹((3x + 1)/2) + C.

Answer(1/6) tan⁻¹((3x + 1)/2) + C

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1. Complete the square: 7 − 6x − x² = 16 − (x + 3)².
2. Let t = x + 3, so the integral is ∫ dt/√(4² − t²).
3. Use ∫dt/√(a²−t²) = sin⁻¹(t/a) + C, with a = 4.
4. This gives sin⁻¹(t/4) + C.
5. Put back t = x + 3: sin⁻¹((x + 3)/4) + C.

Answersin⁻¹((x + 3)/4) + C

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1. Multiply out: (x − 1)(x − 2) = x² − 3x + 2.
2. Complete the square: x² − 3x + 2 = (x − 3/2)² − (1/2)².
3. Let t = x − 3/2, so the integral is ∫ dt/√(t² − (1/2)²).
4. Use ∫dt/√(t²−a²) = log|t + √(t²−a²)| + C, with a = 1/2.
5. Put back t = x − 3/2: log|(x − 3/2) + √((x − 1)(x − 2))| + C.

Answerlog|(x − 3/2) + √((x − 1)(x − 2))| + C

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1. Complete the square: 8 + 3x − x² = 41/4 − (x − 3/2)².
2. Let t = x − 3/2, so the integral is ∫ dt/√((√41/2)² − t²).
3. Use ∫dt/√(a²−t²) = sin⁻¹(t/a) + C, with a = √41/2.
4. This gives sin⁻¹(t ÷ (√41/2)) + C = sin⁻¹(2t/√41) + C.
5. Put back t = x − 3/2: sin⁻¹((2x − 3)/√41) + C.

Answersin⁻¹((2x − 3)/√41) + C

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1. Multiply out: (x − a)(x − b) = x² − (a + b)x + ab.
2. Complete the square: this equals (x − (a+b)/2)² − ((a−b)/2)².
3. Let t = x − (a+b)/2, so the integral is ∫ dt/√(t² − ((a−b)/2)²).
4. Use ∫dt/√(t²−k²) = log|t + √(t²−k²)| + C.
5. Put back t = x − (a+b)/2: log|(x − (a+b)/2) + √((x − a)(x − b))| + C.

Answerlog|(x − (a+b)/2) + √((x − a)(x − b))| + C

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1. Check: the derivative of 2x² + x − 3 is 4x + 1, exactly the numerator.
2. So the integral has the form ∫ f′(x)/√f(x) dx, with f(x) = 2x² + x − 3.
3. This standard shape integrates to 2√f(x) + C.
4. So the integral is 2√(2x² + x − 3) + C.

Answer2√(2x² + x − 3) + C

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1. Split the fraction: (x + 2)/√(x²−1) = x/√(x²−1) + 2/√(x²−1).
2. For the first part, ∫x/√(x²−1) dx = √(x² − 1), since the numerator is half the derivative of x²−1.
3. For the second part, use ∫dx/√(x²−a²) = log|x + √(x²−a²)| + C, with a = 1.
4. This gives 2 log|x + √(x² − 1)| + C.
5. Together: √(x² − 1) + 2 log|x + √(x² − 1)| + C.

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

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1. The derivative of the denominator 3x² + 2x + 1 is 6x + 2. Write 5x − 2 = p(6x + 2) + q.
2. Compare the x terms: 6p = 5, so p = 5/6. Compare the constants: 2p + q = −2, so q = −2 − 5/3 = −11/3.
3. So the integral is (5/6) ∫ (6x + 2)/(3x² + 2x + 1) dx − (11/3) ∫ dx/(3x² + 2x + 1).
4. First part: the numerator is the derivative of the denominator, so it gives (5/6) log|3x² + 2x + 1|. Since 3x² + 2x + 1 is always positive, this is (5/6) log(3x² + 2x + 1).
5. Second part: complete the square, 3x² + 2x + 1 = 3[(x + 1/3)² + 2/9] = 3[(x + 1/3)² + (√2/3)²].
6. So ∫ dx/(3x² + 2x + 1) = (1/3) ∫ dx/((x + 1/3)² + (√2/3)²). Using ∫dt/(t² + a²) = (1/a) tan⁻¹(t/a) with a = √2/3, this is (1/3)·(3/√2)·tan⁻¹((x + 1/3)/(√2/3)) = (1/√2) tan⁻¹((3x + 1)/√2).
7. Multiply by −11/3: −(11/(3√2)) tan⁻¹((3x + 1)/√2) = −(11√2/6) tan⁻¹((3x + 1)/√2).
8. Together: (5/6) log(3x² + 2x + 1) − (11√2/6) tan⁻¹((3x + 1)/√2) + C.

Answer(5/6) log(3x² + 2x + 1) − (11√2/6) tan⁻¹((3x + 1)/√2) + C

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1. Multiply out: (x−5)(x−4) = x² − 9x + 20, whose derivative is 2x − 9.
2. Write 6x + 7 = 3(2x − 9) + 34.
3. The 3(2x−9) part gives 3·2√(x²−9x+20) = 6√(x²−9x+20).
4. For the 34 part, complete the square: x² − 9x + 20 = (x − 9/2)² − (1/2)², giving 34 log|(x−9/2) + √(x²−9x+20)| + C.
5. Together: 6√(x² − 9x + 20) + 34 log|(x − 9/2) + √(x² − 9x + 20)| + C.

Answer6√(x² − 9x + 20) + 34 log|(x − 9/2) + √(x² − 9x + 20)| + C

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1. Write 4x − x² = 4 − (x − 2)², and let t = x − 2, so x + 2 = t + 4.
2. The integral becomes ∫(t + 4)/√(4 − t²) dt = ∫t/√(4−t²) dt + 4∫dt/√(4−t²).
3. ∫t/√(4−t²) dt = −√(4 − t²), since the numerator is minus half the derivative of 4−t².
4. ∫dt/√(4−t²) = sin⁻¹(t/2) + C, using a = 2.
5. Together, and putting back t = x − 2: −√(4x − x²) + 4 sin⁻¹((x − 2)/2) + C.

Answer−√(4x − x²) + 4 sin⁻¹((x − 2)/2) + C

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1. The number under the root is x² + 2x + 3. Its derivative is 2x + 2.
2. Write the top, x + 2, using that: x + 2 = (1/2)(2x + 2) + 1.
3. So the integral splits into (1/2)∫(2x + 2)/√(x² + 2x + 3) dx + ∫1/√(x² + 2x + 3) dx.
4. The first part is a direct power-rule integral: it gives √(x² + 2x + 3).
5. For the second part, complete the square: x² + 2x + 3 = (x + 1)² + 2.
6. ∫dx/√((x + 1)² + 2) is a standard form. It gives log|(x + 1) + √(x² + 2x + 3)|.
7. Add the two results together.

Answer√(x² + 2x + 3) + log|(x + 1) + √(x² + 2x + 3)| + C

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1. The derivative of x² − 2x − 5 is 2x − 2.
2. Write the top, x + 3, using that: x + 3 = (1/2)(2x − 2) + 4.
3. So the integral splits into (1/2)∫(2x − 2)/(x² − 2x − 5) dx + 4∫dx/(x² − 2x − 5).
4. The first part gives (1/2) log|x² − 2x − 5|.
5. For the second part, complete the square: x² − 2x − 5 = (x − 1)² − 6.
6. ∫dx/((x − 1)² − 6) is the standard log form for a difference of squares: (1/(2√6)) log|(x − 1 − √6)/(x − 1 + √6)|.
7. Multiply this by 4, which gives (√6/3) log|(x − 1 − √6)/(x − 1 + √6)|.
8. Add the two results together.

Answer(1/2) log|x² − 2x − 5| + (√6/3) log|(x − 1 − √6)/(x − 1 + √6)| + C

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1. The derivative of x² + 4x + 10 is 2x + 4.
2. Write the top, 5x + 3, using that: 5x + 3 = (5/2)(2x + 4) − 7.
3. So the integral splits into (5/2)∫(2x + 4)/√(x² + 4x + 10) dx − 7∫dx/√(x² + 4x + 10).
4. The first part is a power-rule integral: it gives 5√(x² + 4x + 10).
5. For the second part, complete the square: x² + 4x + 10 = (x + 2)² + 6.
6. ∫dx/√((x + 2)² + 6) is a standard form: log|(x + 2) + √(x² + 4x + 10)|.
7. Multiply this by −7 and add to the first part.

Answer5√(x² + 4x + 10) − 7 log|(x + 2) + √(x² + 4x + 10)| + C

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1. Complete the square in the denominator: x² + 2x + 2 = (x + 1)² + 1.
2. ∫dx/((x + 1)² + 1) is the standard inverse-tangent form, which gives tan⁻¹(x + 1) + C.
3. Checking by differentiating each option, only tan⁻¹(x + 1) + C gives back 1/(x² + 2x + 2).

AnswerOption (B): tan⁻¹(x + 1) + C

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1. Complete the square under the root: 9x − 4x² = 4[(9/8)² − (x − 9/8)²].
2. So √(9x − 4x²) = 2√((9/8)² − (x − 9/8)²).
3. ∫dx/[2√((9/8)² − (x − 9/8)²)] is the standard inverse-sine form: (1/2) sin⁻¹((x − 9/8)/(9/8)) + C.
4. Simplify (x − 9/8)/(9/8) to (8x − 9)/9, giving (1/2) sin⁻¹((8x − 9)/9) + C.
5. Checking by differentiating each option, only this one gives back 1/√(9x − 4x²).

AnswerOption (B): (1/2) sin⁻¹((8x − 9)/9) + C

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