If ∫ dx (x + 1) ^ 2 ( x ^ 2 + 2x + 2) = (A)/(x + 1) + B ^ - 1 (x +...

∫ \dfrac{dx}{{\left(x + 1\right)}^{2}\left({x}^{2} + 2x + 2\right)} = \dfrac{A}{x + 1} + B\tan^{- 1} \left(x + 1\right) + C,

where

A and B

are constants and

C is the constant of integration, then \left|A - B\right|

is equal to

Correct answer

Step-by-step solution

∫ \dfrac{dx}{{\left(x + 1\right)}^{2}\left({x}^{2} + 2x + 2\right)} = ∫ \dfrac{\left({x}^{2} + 2x + 2\right) - \left({x}^{2} + 2x + 1\right)}{\left({x}^{2} + 2x + 1\right)\left({x}^{2} + 2x + 2\right)}dx $ $ = ∫ \dfrac{dx}{{\left(x + 1\right)}^{2}} - ∫ \dfrac{dx}{{\left(x + 1\right)}^{2} + 1} $ $ = - \dfrac{1}{x + 1} - \tan^{- 1} \left(x + 1\right) + C $ So$ A = B = - 1 $Hence, \left|A - B\right| = 0$

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