integral of 6x-5 root of 6-2x^2+xdx evaluate
Answers
Given : (6x - 5)√(6 - 2x² + x)dx
To Find : Integrate
Solution:
(6x - 5)√(6 - 2x² + x)
= (3/2)(4x - 10/3)√(6 - 2x² + x)
= (3/2)(4x -1 + 1 - 10/3)√(6 - 2x² + x)
= (3/2)(4x -1 - 7/3)√(6 - 2x² + x)
= (3/2)(4x -1)√(6 - 2x² + x) - (7/2)√(6 - 2x² + x)
I = I1 - I2
I1 = ∫(3/2)(4x -1)√(6 - 2x² + x) dx
(6 - 2x² + x) = t
=> (-4x + 1)dx = dt
=> (4x - 1)dx = - dt
I1 = -∫(3/2) √ t dt
I1 = - (3/2) t^(3/2) / (3/2) + c1
I1 = - (6 - 2x² + x)^(3/2) + c1
I2 = ∫√2 (7/2)√(3 - x² + x/2)dx
I2 = ∫√2 (7/2)√(3+ 1/16 - 1/16 - x² + x/2)dx
I2 =∫√2 (7/2)√(49/16 - ( (1/4)² + x² -2.x(1/4))dx
I2 = (7/√2)∫√((7/4)² - ( x -1/4)²)dx
x -1/4 = y
dx = dy
I2 = (7/√2)∫√((7/4)² - ( y)²)dy
I2 = (7/√2) { (y/2)√((7/4)² - ( y)² + (7/4)²/2 sin⁻¹(y/(7/4)) + c2
substitute y = x -1/4
I2 = (7/√2) { ((x -1/4)/2)√((7/4)² - ( x -1/4)² + (7/4)²/2 sin⁻¹((x -1/4)/(7/4)) + c2
I2 = (7/√2) { ((4x -1 )/8)√(3 - x² + x/2) + (49/32) sin⁻¹((4x -1)/7) + c2
I2 = (7 (4x -1 )/16)√(6 - 2x² + x ) + (343/32√2) sin⁻¹((4x -1)/7) + c2
I = I1 - I2
= - (6 - 2x² + x)^(3/2) - (7 (4x -1 )/16)√(6 - 2x² + x ) - (343/32√2) sin⁻¹((4x -1)/7) +C
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