Mathematics

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


ANSWER

$$\frac{1}{2}\ln \left| {2{x^2} + 2x + 1} \right| - 2{\tan ^{ - 1}}\left( {2x + 1} \right) + c$$


SOLUTION
$$\begin{array}{l} \frac { 1 }{ 2 } \left[ { \int { \frac { { 2x+1 } }{ { { x^{ 2 } }+x+\frac { 1 }{ 2 }  } } dx-\int { \frac { 2 }{ { { x^{ 2 } }+x+\frac { 1 }{ 2 }  } }  }  } dx } \right]  \\ { x^{ 2 } }+x+\frac { 1 }{ 2 } =t,\, \, \, \, x+\frac { 1 }{ 2 } =u={ x^{ 2 } }+x+\frac { 1 }{ 4 } ={ u^{ 2 } } \\ \left( { 2x+1 } \right) dx=dt,\, \, \, \, x+\frac { 1 }{ 2 } =u \\ \frac { 1 }{ 2 } \left[ { \int { \frac { { dt } }{ t } -2\int { \frac { 1 }{ { { u^{ 2 } }+\frac { 1 }{ u }  } } du }  }  } \right]  \\ =\frac { 1 }{ 2 } \left[ { \ln { \left| t \right|  } -2.\left( { \frac { 1 }{ { \frac { 1 }{ 2 }  } }  } \right) { { \tan   }^{ -1 } }\left( { \frac { u }{ { \frac { 1 }{ 2 }  } }  } \right)  } \right] +c \\ =\frac { 1 }{ 2 } \ln { \left| t \right|  } -2{ \tan ^{ -1 }  }2u+c \\ =\frac { 1 }{ 2 } \ln { \left| { { x^{ 2 } }+x+\frac { 1 }{ 2 }  } \right|  } -2{ \tan ^{ -1 }  }\left( { 2x+1 } \right) +c \\ =\frac { 1 }{ 2 } \ln { \left| { 2{ x^{ 2 } }+2x+1 } \right|  } -2{ \tan ^{ -1 }  }\left( { 2x-1 } \right) +c \\ =\frac { 1 }{ 2 } \ln { \left| { 2{ x^{ 2 } }+2x+1 } \right|  } -2{ \tan ^{ -1 }  }\left( { 2x+1 } \right) +{ c_{ 1 } } \\ { c_{ 1 } }=c-\frac { 1 }{ 2 } \ln { 2 }  \end{array}$$
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Single Correct Medium Published on 17th 09, 2020
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