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Question

Primitive of $$\displaystyle \dfrac{3x^4-1}{(x^4+x+1)^2}$$ w.r.t  $$x$$ is


A
x(x4+x+1)+c
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B
x(x4+x+1)+c
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C
x+1(x4+x+1)+c
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D
x+1(x4+x+1)+c
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Solution

The correct option is B $$-\displaystyle \dfrac{x}{(x^4+x+1)}+c$$
$$\displaystyle \int { \dfrac { { 3x }^{ 4 }-1 }{ { \left( { x }^{ 4 }+x+1 \right)  }^{ 2 } }  } dx=\int { \dfrac { { 3x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } }  }{ { \left( { x }^{ 3 }+1+\dfrac { 1 }{ x }  \right)  }^{ 2 } }  } dx$$
Substitute $$\displaystyle { x }^{ 3 }+1+\dfrac { 1 }{ x } =t\Rightarrow \left( { 3x }^{ 2 }-\dfrac { 1 }{ { x }^{ 2 } }  \right) dx=dt$$
$$\displaystyle \therefore \int { \dfrac { dt }{ { t }^{ 2 } }  } =-\dfrac { 1 }{ t } +c=-\dfrac { 1 }{ { x }^{ 3 }+1+\dfrac { 1 }{ x }  } +c$$
$$\displaystyle =-\dfrac { x }{ { x }^{ 4 }+x+1 } +c$$

Mathematics

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