Definite Integration
Integration by substitution
Grade Class 12
Question:
If ∫ \frac{(x-1) dx}{x^2 \sqrt{2x^2 - 2x + 1}} = \frac{\sqrt{f(x)}}{g(x)} + C, where f(x) is a quadratic expression and g(x) is a monic linear expression.
(A) f(x) = 2x^2 - 2x + 1
(B) g(x) = x + 1
(C) g(x) = x
(D) f(x) = 2x^2 - 2x
Step-by-Step Solution
Key Concept: Substitute x = 1/t to transform the integral into a standard form.
Let x = 1/t, then dx = -1/t^2 dt. The integral becomes \int (1/t - 1) (-1/t^2) dt / ((1/t^2) * sqrt(2/t^2 - 2/t + 1)) = \int (1-t)/t * dt / (sqrt(2-2t+t^2)/t) = \int (1-t) dt / sqrt(t^2-2t+2) = - \int (t-1) dt / sqrt((t-1)^2 + 1). Let u = t-1, du = dt. Integral = - \int u du / sqrt(u^2+1) = -sqrt(u^2+1) + C = -sqrt((1/x - 1)^2 + 1) + C = -sqrt((1-2x+x^2+x^2)/x^2) + C = -sqrt(2x^2-2x+1)/x + C. Comparing with sqrt(f(x))/g(x), we get f(x) = 2x^2-2x+1 and g(x) = x.
Correct Answer: A,C