Mathematics

Solve:
$$\displaystyle \int \dfrac{1}{(x+1)(x^2+1)}dx$$


SOLUTION
Given the integral,
$$\int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) }  } dx$$
using partial fraction we get,
$$\int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) }  } dx\\ =\int { [\dfrac { 1 }{ 2(x+1) } -\dfrac { x-1 }{ 2({ x }^{ 2 }+1) } ] } dx\\ =\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ x+1 }  } dx-\dfrac { 1 }{ 2 } \int { \dfrac { x-1 }{ { x }^{ 2 }+1 }  } dx$$
Now, for $$\int { \dfrac { 1 }{ x+1 }  } dx$$
Let, $$u=x+1\\ \Rightarrow \dfrac { du }{ dx } =1\\ \Rightarrow du=dx$$
Substituting the values of $$u$$ and $$du$$ we get,
$$\int { \dfrac { 1 }{ x+1 }  } dx\\ =\int { \dfrac { 1 }{ u }  } du\\ =\ln { (u) } \\ =\ln { (x+1) } $$
for $$\int { \dfrac { x-1 }{ { x }^{ 2 }+1 }  } dx$$
Expanding the integral,
$$\int { \dfrac { x-1 }{ { x }^{ 2 }+1 }  } dx\\ =\int { \dfrac { x }{ { x }^{ 2 }+1 }  } dx-\int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx$$
Again, for $$\int { \dfrac { x }{ { x }^{ 2 }+1 }  } dx$$
Let, 
$$u={ x }^{ 2 }+1\\ \Rightarrow \dfrac { du }{ dx } =2x\\ \Rightarrow dx=\dfrac { 1 }{ 2x } du\\ \therefore \int { \dfrac { x }{ { x }^{ 2 }+1 }  } dx\\ =\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ u }  } du\\ =\dfrac { \ln { (u) }  }{ 2 } \\ =\dfrac { \ln { ({ x }^{ 2 }+1) }  }{ 2 } $$
And 
$$\int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx\\ =\arctan { (x) } \\ \therefore \int { \dfrac { x }{ { x }^{ 2 }+1 }  } dx-\int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx\\ =\dfrac { \ln { ({ x }^{ 2 }+1) }  }{ 2 } -\arctan { (x) } $$
So,
$$\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ x+1 }  } dx-\dfrac { 1 }{ 2 } \int { \dfrac { x-1 }{ { x }^{ 2 }+1 }  } dx\\ =\dfrac { \ln { (x+1) }  }{ 2 } -\dfrac { \ln { ({ x }^{ 2 }+1) }  }{ 4 } +\dfrac { \arctan { (x) }  }{ 2 } \\ \therefore \int { \dfrac { 1 }{ (x+1)({ x }^{ 2 }+1) }  } dx=\dfrac { \ln { (x+1) }  }{ 2 } -\dfrac { \ln { ({ x }^{ 2 }+1) }  }{ 4 } +\dfrac { \arctan { (x) }  }{ 2 } +C.$$ 
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Subjective Medium Published on 17th 09, 2020
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