CBSE Class 12-science Answered

Put tan^-1(2x/(1 - x²)) = y

dy/dx = (1/(1 + (2x/(1 - x²))²) [(1 - x²).2 - 2x (-2x)]/(1 - x²)²

{Note the use of chain rule and standard result d(tan^-1x)/dx = 1/(1 + x²)}

Simplifying gives,

dy/dx = 2/(1+x²)...(1)

2x/(1-x²) = tan y

tan y x² +2x - tan y = 0

x = (-1±sec y)/tan y.

1 + x² = 2sec y(sec y ± 1)/tan² y = 2(cosec²y ± cosec²y cos y)

2(cosec²y ± cosec²y cos y) dy = dx.

tan^-1(2x/(1 - x²)) dx = 2y(cosec²y ± cosec²y cos y) dy

This can be separated and integrated by parts.

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