Definite Integration
Integration
Grade Class 12
Question:
If ∫ <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mo>(</mo><mi>x</mi><mo>-</mo><mn>1</mn><mo>)</mo><mo> </mo><mi>d</mi><mi>x</mi></mrow><mrow><msup><mi>x</mi><mn>2</mn></msup><msqrt><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>-</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>1</mn></msqrt></mrow></mfrac></math> = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msqrt><mi>f</mi><mo>(</mo><mi>x</mi><mo>)</mo></msqrt><mrow><mi>g</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mfrac></math> + C, where f(x) is a quadratic expression and g(x) is a monic linear expression.
(A) f(x) = 2x<sup>2</sup> - 2x + 1
(B) g(x) = x + 1
(C) g(x) = x
(D) f(x) = 2x<sup>2</sup> - 2x
Step-by-Step Solution
Key Concept: Substitute t = 1/x, then the integral becomes \int -dt / sqrt(2 - 2t + t^2). Complete the square in the denominator to solve.
Let t = 1/x, then dt = -1/x^2 dx. The integral becomes \int -dt / sqrt(2 - 2t + t^2) = -\int dt / sqrt((t-1)^2 + 1) = -ln|t-1 + sqrt((t-1)^2 + 1)| + C. Alternatively, rewrite the integral as \int (1/x - 1/x^2) dx / sqrt(2 - 2/x + 1/x^2). Let u = 1/x, du = -1/x^2 dx. The expression simplifies to 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: 1, 3