Mathematics

# Evaluate $\int \sqrt{1+t^{3}} dt$

##### SOLUTION
We have,
$I=\int _{ }^{ }{ \sqrt { 1+{ t^{ 3 } } } } dt \\I ={ \int _{ }^{ }{ \left( { 1+{ t^{ 3 } } } \right) } ^{ 1/2 } }dt \\I =\int _{ }^{ }{ \left( { 1+\dfrac { 1 }{ 2 } { t^{ 3 } }-\dfrac { 1 }{ 8 } { t^{ 6 } }....... } \right) } dt \\I =\left( { t+\dfrac { 1 }{ 8 } { t^{ 4 } }-\dfrac { 1 }{ { 56 } } .{ t^{ 7 } }+...... } \right) +C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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