Exercise 7.7 answers: Integrals

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1. This has the shape √(a² − x²) with a = 2.
2. The standard result is ∫√(a²−x²)dx = (x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a) + C.
3. Put a = 2: (x/2)√(4−x²) + (4/2)sin⁻¹(x/2) + C.
4. So the integral is (x/2)√(4−x²) + 2sin⁻¹(x/2) + C.

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

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1. Write 1 − 4x² as 1 − (2x)², a √(a²−t²) shape with t = 2x, a = 1.
2. Put t = 2x, so dt = 2dx, i.e. dx = dt/2.
3. ∫√(1−t²)(dt/2) = (1/2)[(t/2)√(1−t²) + (1/2)sin⁻¹t] + C, using the standard result with a=1.
4. Put t back as 2x: (1/2)[x√(1−4x²) + (1/2)sin⁻¹(2x)] + C.
5. So the integral is (x/2)√(1−4x²) + (1/4)sin⁻¹(2x) + C.

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

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1. Complete the square: x² + 4x + 6 = (x+2)² + 2.
2. This is the √(y²+a²) shape with y = x+2, a² = 2.
3. Standard result: ∫√(y²+a²)dy = (y/2)√(y²+a²) + (a²/2)log|y+√(y²+a²)| + C.
4. Substitute y = x+2, a² = 2: ((x+2)/2)√(x²+4x+6) + 1·log|x+2+√(x²+4x+6)| + C.
5. So the integral is ((x+2)/2)√(x²+4x+6) + log|x+2+√(x²+4x+6)| + C.

Answer((x+2)⁄2)√(x² + 4x + 6) + log|x + 2 + √(x² + 4x + 6)| + C

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1. Complete the square: x² + 4x + 1 = (x+2)² − 3.
2. This is the √(y²−a²) shape with y = x+2, a² = 3.
3. Standard result: ∫√(y²−a²)dy = (y/2)√(y²−a²) − (a²/2)log|y+√(y²−a²)| + C.
4. Substitute y = x+2, a² = 3: ((x+2)/2)√(x²+4x+1) − (3/2)log|x+2+√(x²+4x+1)| + C.

Answer((x+2)⁄2)√(x² + 4x + 1) − (3⁄2) log|x + 2 + √(x² + 4x + 1)| + C

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1. Write 1 − 4x − x² = −(x²+4x−1) = −[(x+2)²−5] = 5 − (x+2)².
2. This is the √(a²−y²) shape with y = x+2, a² = 5.
3. Standard result: ∫√(a²−y²)dy = (y/2)√(a²−y²) + (a²/2)sin⁻¹(y/a) + C.
4. Substitute y = x+2, a² = 5, a = √5: ((x+2)/2)√(1−4x−x²) + (5/2)sin⁻¹((x+2)/√5) + C.

Answer((x+2)⁄2)√(1 − 4x − x²) + (5⁄2) sin⁻¹((x+2)⁄√5) + C

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1. Complete the square: x² + 4x − 5 = (x+2)² − 9.
2. This is the √(y²−a²) shape with y = x+2, a² = 9.
3. Standard result: ∫√(y²−a²)dy = (y/2)√(y²−a²) − (a²/2)log|y+√(y²−a²)| + C.
4. Substitute y = x+2, a² = 9: ((x+2)/2)√(x²+4x−5) − (9/2)log|x+2+√(x²+4x−5)| + C.

Answer((x+2)⁄2)√(x² + 4x − 5) − (9⁄2) log|x + 2 + √(x² + 4x − 5)| + C

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1. Write 1 + 3x − x² = −(x² − 3x − 1) = −[(x − 3/2)² − 13/4] = 13/4 − (x − 3/2)².
2. This is the √(a²−y²) shape with y = x − 3/2, a² = 13/4.
3. Standard result: ∫√(a²−y²)dy = (y/2)√(a²−y²) + (a²/2)sin⁻¹(y/a) + C.
4. Substitute y = x−3/2, a = √13/2: y/2 = (2x−3)/4, and y/a = (2x−3)/√13.
5. So the integral is ((2x−3)/4)√(1+3x−x²) + (13/8)sin⁻¹((2x−3)/√13) + C.

Answer((2x − 3)⁄4)√(1 + 3x − x²) + (13⁄8) sin⁻¹((2x − 3)⁄√13) + C

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1. Complete the square: x² + 3x = (x + 3/2)² − 9/4.
2. This is the √(y²−a²) shape with y = x + 3/2, a² = 9/4.
3. Standard result: ∫√(y²−a²)dy = (y/2)√(y²−a²) − (a²/2)log|y+√(y²−a²)| + C.
4. Substitute y = x+3/2, a² = 9/4: y/2 = (2x+3)/4, a²/2 = 9/8.
5. So the integral is ((2x+3)/4)√(x²+3x) − (9/8)log|x+3/2+√(x²+3x)| + C.

Answer((2x + 3)⁄4)√(x² + 3x) − (9⁄8) log|x + 3⁄2 + √(x² + 3x)| + C

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1. Write 1 + x²/9 = (9+x²)/9, so √(1+x²/9) = (1/3)√(x²+9).
2. This is (1/3) times the √(y²+a²) shape with a = 3.
3. Standard result: ∫√(x²+a²)dx = (x/2)√(x²+a²) + (a²/2)log|x+√(x²+a²)| + C.
4. With a=3: (x/2)√(x²+9) + (9/2)log|x+√(x²+9)| + C, then multiply by 1/3.
5. So the integral is (x/6)√(x²+9) + (3/2)log|x+√(x²+9)| + C.

Answer(x⁄6)√(x² + 9) + (3⁄2) log|x + √(x² + 9)| + C

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1. This is the standard result √(x²+a²) with a = 1.
2. ∫√(x²+a²)dx = (x/2)√(x²+a²) + (a²/2)log|x+√(x²+a²)| + C.
3. With a=1: (x/2)√(1+x²) + (1/2)log|x+√(1+x²)| + C.
4. This matches option (A).

Answer(A) (x⁄2)√(1+x²) + (1⁄2) log(x+√(1+x²)) + C

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1. Complete the square: x² − 8x + 7 = (x − 4)² − 9.
2. This is the √(y² − a²) shape with y = x − 4 and a² = 9.
3. Standard result: ∫√(y² − a²) dy = (y⁄2)√(y² − a²) − (a²⁄2) log|y + √(y² − a²)| + C.
4. With y = x − 4 and a²⁄2 = 9⁄2: (1⁄2)(x − 4)√(x² − 8x + 7) − (9⁄2) log|x − 4 + √(x² − 8x + 7)| + C.
5. This is option (D). Option (A) has +9 before the log, option (B) has x + 4 and +9, and option (C) has −3√2 — each differs from this result.

Answer(D) (1⁄2)(x − 4)√(x² − 8x + 7) − (9⁄2) log|x − 4 + √(x² − 8x + 7)| + C

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