Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 years ago

# How to find ∫ 1/√(x -1)(3 - x) dx?

(x - 1)(3 - x) is all under the root

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• Anonymous
10 years ago

Expand (x - 1)(3 - x) to get:

∫ 1/√[(x - 1)(3 - x)] dx

= ∫ 1/√(3x - x^2 - 3 + x) dx

= ∫ 1/√(-x^2 + 4x - 3) dx.

Then, by completing the square:

∫ 1/√(-x^2 + 4x - 3) dx

= ∫ 1/√[-(x^2 - 4x + 3)] dx

= ∫ 1/√[1 - (x^2 - 4x + 4)] dx

= ∫ 1/√[1 - (x - 2)^2] dx.

Next, letting u = x - 2 ==> du = dx yields:

∫ 1/√[1 - (x - 2)^2] dx

= ∫ 1/√(1 - u^2) du, by substituting du in for x - 2 and du in for dx

= arcsin(u) + C, ∫ 1/√(1 - x^2) dx = arcsin(x) + C

= arcsin(x - 2) + C, since u = x - 2.

I hope this helps!

• 10 years ago

1) Expanding the function withing the square root, it is ∫ dx/√(4x - 3 - x²)

2) ==> ∫ dx/√(4x - 3 - x²) = ∫ dx/√(4x - 4 + 1 - x²) = ∫ dx/√{1 - (x-2)²}

3) Let x-2 = t; ==> dx = dt

4) Hence, it is ∫ dt/√(1 - t²) = sin⁻¹t

Substituting back for t, the answer is: 'sin⁻¹(x - 2) + C'