Mathematics

Solve:
$$\displaystyle \int \dfrac{x^6+1}{x^2+1}\ dx$$


SOLUTION
$$\displaystyle\int { \dfrac { { x }^{ 6 }+1 }{ { x }^{ 2}+1 } dx } =$$ $$I$$

Substitute $$x=\tan t$$,$$dx=$$$${ sec }^{ 2 }tdt$$

$$\therefore \quad I=\displaystyle\int { \left( \dfrac { { tan }^{ 6 }t+1 }{ { tan }^{ 2 }t+1 }  \right)  } { sec }^{ 2 }tdt$$

$$\therefore \quad I=\int { \left( { tan }^{ 6 }t+1 \right)  } dt$$

Now $$\therefore \int { { tan }^{ 6 } } dt=\int { { tan }^{ 4 }t } { sec }^{ 2 }tdt-\int { { tan }^{ 4 } } dt$$

$$=\dfrac { { tan }^{ 5 }t }{ 5 } -\dfrac { { tan }^{ 3 }t }{ 3 } +\int { \left( { sec }^{ 2 }t-1 \right) dt } $$

$$\int { { tan }^{ 6 } } dt=\dfrac { { tan }^{ 5 }t }{ 5 } -\frac { { tan }^{ 3 }t }{ 3 } +tant-t+c$$

Substituting we get

$$\displaystyle\int { \dfrac { { x }^{ 6 }+41 }{ { x }^{ 2 }+1 }  } dx=t+\dfrac { { tan }^{ 5 }t }{ 5 } -\dfrac { { tan }^{ 3 }t }{ 3 } +tant-t+c$$

$$\displaystyle\int { \dfrac { { x }^{ 6 }+1 }{ { x }^{ 2 }+1 }  } dx=\dfrac { { x }^{ 5 } }{ 5 } -\dfrac { { x }^{ 3 } }{ 3 } +x+c$$

Ans $$\displaystyle\int { \dfrac { { x }^{ 6 }+1 }{ { x }^{ 2 }+1 }  } dx=\dfrac { { x }^{ 5 } }{ 5 } -\dfrac { { x }^{ 3 } }{ 3 } +x+c$$
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Subjective Medium Published on 17th 09, 2020
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