J's answer to Yos's Secondary 4 A Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Integration by substitution (twice)
① let y = √x
Then y² = x → 2y = dx/dy →"dx = 2y dy"
∫ √x √(1 + √x) dx = ∫ y √(1 + y) 2y dy = 2 ∫ y² √(1 + y) dy
② let u = √(1 + y)
Then u² = 1 + y → u² - 1 = y → 2u = dy/du →"dy = 2u du"
Then 2 ∫ y² √(1 + y) dy = 2 ∫ (u² - 1)² u (2u) du
= 4 ∫ (u⁴ - 2u² + 1)u² du
= 4 ∫ (u⁶ - 2u⁴ + u²) du
= 4 (1/7 u⁷ - 2/5 u⁵ + ⅓ u³)
= 4/7 (1 + y)⁷/² - 8/5 (1 + y)⁵/² + 4/3 (1 + y)³/²
= 4/7 (1 + √x)⁷/² - 8/5 (1 + √x)⁵/² + 4/3 (1 + √x)³/² + c
You can choose to factorise or rearrange the terms from here.
① let y = √x
Then y² = x → 2y = dx/dy →"dx = 2y dy"
∫ √x √(1 + √x) dx = ∫ y √(1 + y) 2y dy = 2 ∫ y² √(1 + y) dy
② let u = √(1 + y)
Then u² = 1 + y → u² - 1 = y → 2u = dy/du →"dy = 2u du"
Then 2 ∫ y² √(1 + y) dy = 2 ∫ (u² - 1)² u (2u) du
= 4 ∫ (u⁴ - 2u² + 1)u² du
= 4 ∫ (u⁶ - 2u⁴ + u²) du
= 4 (1/7 u⁷ - 2/5 u⁵ + ⅓ u³)
= 4/7 (1 + y)⁷/² - 8/5 (1 + y)⁵/² + 4/3 (1 + y)³/²
= 4/7 (1 + √x)⁷/² - 8/5 (1 + √x)⁵/² + 4/3 (1 + √x)³/² + c
You can choose to factorise or rearrange the terms from here.
Date Posted:
3 years ago
These are some of the alternative expressions that are equivalent when you rearrange :
4/3 x (√x + 1)³/² - 16/15 √x (√x + 1)⁵/² + 32/105 (√x + 1)⁷/² + c
4/105 (√x + 1)³/² (15x - 12 √x + 8) + c
4/3 x (√x + 1)³/² - 16/15 √x (√x + 1)⁵/² + 32/105 (√x + 1)⁷/² + c
4/105 (√x + 1)³/² (15x - 12 √x + 8) + c
360 Tutoring Program