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Integration of 2^1/x/x^2 |
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Answer» Answer:-ln(|X|)+ln(|x+1|)+ln(|x-1|)+C Step-by-step explanation: LET's factorise the denominator x ( x 2 − 1 ) = x ( x + 1 ) ( x − 1 ) So, let's do the decomposition into partial fractions x 2 + 1 x ( x 2 − 1 ) = x 2 + 1 x ( x + 1 ) ( x − 1 ) = A x + x + 1 + C x − 1 = A ( x + 1 ) ( x − 1 ) + B ( x ) ( x − 1 ) + C ( x ) ( x + 1 ) x ( x + 1 ) ( x − 1 ) The DENOMINATORS are the same, so, we can equalise the numerators x 2 + 1 = A ( x + 1 ) ( x − 1 ) + B ( x ) ( x − 1 ) + C ( x ) ( x + 1 ) Let x = 0 , ⇒ , 1 = − A , ⇒ , A = − 1 Let x = − 1 , ⇒ , 2 = 2 B , ⇒ , B = 1 Let x = 1 , ⇒ , 2 = 2 C , ⇒ , C = 1 Therefore, x 2 + 1 x ( x 2 − 1 ) = − 1 x + 1 x + 1 + 1 x − 1 So, we can do the integration ∫ ( x 2 + 1 ) d x x ( x 2 − 1 ) = − ∫ d x x + ∫ d x x + 1 + ∫ d x x − 1 = − ln ( | x | ) + ln ( | x + 1 | ) + ln ( | x − 1 | ) + C
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