Definite Integration
Integration
Grade Class 12
Question:
The value of ∫ <sup>ln(&frac{x-1}{x+1})</sup>⁄<sub>x<sup>2</sup> - 1</sub> dx is equal to
(A) &frac{1}{2} ln<sup>2</sup> (&frac{x-1}{x+1}) + C
(B) &frac{1}{4} ln<sup>2</sup> (&frac{x-1}{x+1}) + C
(C) &frac{1}{2} ln<sup>2</sup> (&frac{x+1}{x-1}) + C
(D) &frac{1}{4} ln<sup>2</sup> (&frac{x+1}{x-1}) + C
Step-by-Step Solution
Key Concept: Let t = ln((x-1)/(x+1)). Then dt = (d/dx [ln(x-1) - ln(x+1)]) dx = (1/(x-1) - 1/(x+1)) dx = (2/(x^2-1)) dx. Thus, the integral becomes 1/2 \int t dt = 1/4 t^2 + C.
Let I = ∫ <sup>ln(&frac{x-1}{x+1})</sup>⁄<sub>x<sup>2</sup> - 1</sub> dx. Let t = ln(&frac{x-1}{x+1}). Then dt = (&frac{1}{x-1} - &frac{1}{x+1}) dx = &frac{2}{x<sup>2</sup>-1} dx. So, &frac{dx}{x<sup>2</sup>-1} = &frac{1}{2} dt. The integral becomes ∫ t (&frac{1}{2} dt) = &frac{1}{4} t<sup>2</sup> + C = &frac{1}{4} [ln(&frac{x-1}{x+1})]<sup>2</sup> + C. Since [ln(&frac{x-1}{x+1})]<sup>2</sup> = [ln(&frac{x+1}{x-1})]<sup>2</sup>, both (B) and (D) are correct.
Correct Answer: 1, 4