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

∫dx2x2+x+1=\int_{}^{}\frac{dx}{2x^{2} + x + 1} =

A

17tan⁡−1(4x+17)+c\frac{1}{\sqrt{7}}\tan^{- 1}\left( \frac{4x + 1}{\sqrt{7}} \right) + c

B

127tan⁡−1(4x+17)+c\frac{1}{2\sqrt{7}}\tan^{- 1}\left( \frac{4x + 1}{\sqrt{7}} \right) + c

C

12tan⁡−1(4x+17)+c\frac{1}{2}\tan^{- 1}\left( \frac{4x + 1}{\sqrt{7}} \right) + c

D

None of these

Answer

None of these

Explanation

Solution

I=12∫dxx2+x2+12\mathbf{I =}\frac{\mathbf{1}}{\mathbf{2}}\int_{}^{}\frac{\mathbf{dx}}{\mathbf{x}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}} ⇒I=12∫dx(x+14)2+(74)2\mathbf{\Rightarrow}\mathbf{I =}\frac{\mathbf{1}}{\mathbf{2}}\int_{}^{}\frac{\mathbf{dx}}{\left( \mathbf{x +}\frac{\mathbf{1}}{\mathbf{4}} \right)^{\mathbf{2}}\mathbf{+}\left( \frac{\sqrt{\mathbf{7}}}{\mathbf{4}} \right)^{\mathbf{2}}}

I=12.47.tan⁡−1[x+14(7/4)]+c\mathbf{I =}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{.}\frac{\mathbf{4}}{\sqrt{\mathbf{7}}}\mathbf{.}\mathbf{\tan}^{\mathbf{- 1}}\left\lbrack \frac{\mathbf{x +}\frac{\mathbf{1}}{\mathbf{4}}}{\mathbf{(}\sqrt{\mathbf{7}}\mathbf{/4)}} \right\rbrack\mathbf{+ c} ⇒\mathbf{\Rightarrow} I=27tan⁡−1[4x+17]+c\mathbf{I =}\frac{\mathbf{2}}{\sqrt{\mathbf{7}}}\mathbf{\tan}^{\mathbf{-}\mathbf{1}}\left\lbrack \frac{\mathbf{4x + 1}}{\sqrt{\mathbf{7}}} \right\rbrack\mathbf{+ c}