Mathematics

Evaluate : $$\displaystyle \int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 }  } dx$$


SOLUTION
Let  $$I=\displaystyle\int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 }  } dx$$

$$=\dfrac{1}{2}\displaystyle\int { \dfrac { 1 }{ { x }^{ 2 }+\dfrac{x}{2}+\dfrac{1}{2} }  } dx$$

$$=\dfrac{1}{2}\displaystyle\int { \dfrac { 1 }{ { x }^{ 2 }+\dfrac{x}{2}+\dfrac{1}{16}-\dfrac{1}{16}+\dfrac{1}{2} }  } dx$$

$$=\dfrac{1}{2}\displaystyle\int { \dfrac {1} {{\left( {x+\dfrac{1}{4}}\right)^{2}}-\dfrac{1}{16}+\dfrac{1}{2}}}dx$$

$$=\dfrac{1}{2}\displaystyle\int { \dfrac {1} {{\left( {x+\dfrac{1}{4}}\right)^{2}}+\dfrac{7}{16}}}dx$$

We know that,  $$\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\dfrac{1}{a}\tan^{-1}\dfrac xa+C$$


$$=\dfrac{1}{2}\left(\dfrac{1}{{\dfrac{\sqrt7}{4}}}{tan}^{-1}\left[\dfrac{x+\dfrac{1}{4}}{\dfrac{\sqrt7}{4}}\right]\right)+c$$

$$I=\dfrac{2}{\sqrt7}{tan}^{-1}\left[\dfrac{(4x+1)}{\sqrt7}\right]+c$$

Hence,

$$\displaystyle\int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 }  } dx$$$$=\dfrac{2}{\sqrt7}{tan}^{-1}\left[\dfrac{(4x+1)}{\sqrt7}\right]+c$$

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Subjective Medium Published on 17th 09, 2020
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