Mathematics

$$\int {\frac{{\left( {{x^2} + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} $$


SOLUTION
$$\begin{array}{l} On\, divinding\, \left( { { x^{ 2 } }+1 } \right) by\, \, \left( { { x^{ 2 } }+2x+1 } \right) ,\, we\, get \\ \frac { { \left( { { x^{ 2 } }+1 } \right)  } }{ { { { \left( { x+1 } \right)  }^{ 2 } } } } =\left\{ { 1-\frac { { 2x } }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  } \right\}  \\ Let\, \frac { { 2x } }{ { { { \left( { x+1 } \right)  }^{ 2 } } } } =\frac { A }{ { \left( { x+1 } \right)  } } +\frac { B }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  \\ 2x=A\left( { x+1 } \right) +B.\, \, \, \, ----\left( 1 \right)  \\ On\, equating\, the\, coefficients\, of\, x,\, we\, get\, A=2 \\ On\, equating\, constant\, term\, ,\, we\, get\, \, A+B=0 \\ \Rightarrow B=-A=-2 \\ \therefore \frac { { 2x } }{ { { { \left( { x+1 } \right)  }^{ 2 } } } } =\frac { 2 }{ { \left( { x+1 } \right)  } } -\frac { 2 }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  \\ \therefore I=\int { \left\{ { 1-\frac { { 2x } }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  } \right\}  }  \\ =\int { \left\{ { 1-\frac { 2 }{ { \left( { x+1 } \right)  } } +\frac { 2 }{ { { { \left( { x+1 } \right)  }^{ 2 } } } }  } \right\} dx }  \\ x-2\log  \left| { x+1 } \right| -\frac { 2 }{ { \left( { x+1 } \right)  } } +C. \end{array}$$
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Subjective Medium Published on 17th 09, 2020
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