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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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1 Verified Answer | Published on 17th 09, 2020

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