Evaluate: $\int \frac{2}{(1 - x) (1 + x^{2})} dx$

Solution

$Let \frac{2}{(1 - x) (1 + x^{2})} = \frac{A}{1 - x} + \frac{B x + C}{1 + x^{2}} 2 = A (1 + x^{2}) + (B x + C) (1 - x) 2 = A + A x^{2} + B x - B x^{2} + C - C x$

Equating the coefficients of x², x and constant term, we obtain

A - B = 0

B - C = 0

A + C = 2

On solving these equations, we obtain

A = 1, B = 1, and C = 1

$∴ \frac{2}{(1 - x) (1 + x^{2})} = \frac{1}{1 - x} + \frac{x + 1}{1 + x^{2}} \Rightarrow \int \frac{2}{(1 - x) (1 + x^{2})} dx = \int \frac{1}{1 - x} dx + \int \frac{x}{1 + x^{2}} dx + \int \frac{1}{1 + x^{2}} dx = - \int \frac{1}{x - 1} dx + \frac{1}{2} \int \frac{2 x}{1 + x^{2}} dx + \int \frac{1}{1 + x^{2}} dx = - \log |x - 1| + \frac{1}{2} \log |1 + x^{2}| + \tan^{- 1} x + C$

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Evaluate: $\int \frac{2}{(1 - x) (1 + x^{2})} dx$

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Vector Algebra

Evaluate: ∫ 2( 1 - x ) ( 1 + x2 ) dx

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Evaluate: $\int \frac{2}{(1 - x) (1 + x^{2})} dx$