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Question: \[\int_{}^{}{\frac{x^{2} - 1}{x^{4} + x^{2} + 1}dx =}\]...

∫x2−1x4+x2+1dx=\int_{}^{}{\frac{x^{2} - 1}{x^{4} + x^{2} + 1}dx =}

A

12log⁡[x2+x+1x2−x+1]+c\frac{1}{2}\log\left\lbrack \frac{x^{2} + x + 1}{x^{2} - x + 1} \right\rbrack + c

B

12log⁡[x2−x−1x2+x+1]+c\frac{1}{2}\log\left\lbrack \frac{x^{2} - x - 1}{x^{2} + x + 1} \right\rbrack + c

C

log⁡[x2−x+1x2+x+1]+c\log\left\lbrack \frac{x^{2} - x + 1}{x^{2} + x + 1} \right\rbrack + c

D

12log⁡[x2−x+1x2+x+1]+c\frac{1}{2}\log\left\lbrack \frac{x^{2} - x + 1}{x^{2} + x + 1} \right\rbrack + c

Answer

12log⁡[x2−x+1x2+x+1]+c\frac{1}{2}\log\left\lbrack \frac{x^{2} - x + 1}{x^{2} + x + 1} \right\rbrack + c

Explanation

Solution

∫1−1x2x2+1+1x2dx\int_{}^{}{\frac{\mathbf{1 -}\frac{\mathbf{1}}{\mathbf{x}^{\mathbf{2}}}}{\mathbf{x}^{\mathbf{2}}\mathbf{+ 1 +}\frac{\mathbf{1}}{\mathbf{x}^{\mathbf{2}}}}\mathbf{dx}} =∫1−1x2(x+1x)2−1dx=12log⁡∣x+1x−1x+1x+1∣+c\mathbf{=}\int_{}^{}\frac{\mathbf{1 -}\frac{\mathbf{1}}{\mathbf{x}^{\mathbf{2}}}}{\left( \mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}} \right)^{\mathbf{2}}\mathbf{- 1}}\mathbf{dx =}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\log}\left| \frac{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{- 1}}{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+ 1}} \right|\mathbf{+ c} =12log⁡∣x2+1−xx2+1+x∣+c\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\log}\left| \frac{\mathbf{x}^{\mathbf{2}}\mathbf{+ 1 - x}}{\mathbf{x}^{\mathbf{2}}\mathbf{+ 1 + x}} \right|\mathbf{+ c} =12log⁡∣x2−x+1x2+x+1∣+c\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\log}\left| \frac{\mathbf{x}^{\mathbf{2}}\mathbf{-}\mathbf{x + 1}}{\mathbf{x}^{\mathbf{2}}\mathbf{+ x + 1}} \right|\mathbf{+ c}