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Integral of √(9t^4 + 4t^2)

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Answer to a math question Integral of √(9t^4 + 4t^2)

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$\int{ t\sqrt{ 9{t}^{2}+4 } } \mathrm{d} t$

$\int{ \frac{ 1 }{ 18 } \times \sqrt{ u } } \mathrm{d} u$

$\frac{ 1 }{ 18 } \times \int{ \sqrt{ u } } \mathrm{d} u$

$\frac{ 1 }{ 18 } \times \int{ {u}^{\frac{ 1 }{ 2 }} } \mathrm{d} u$

$\frac{ 1 }{ 18 } \times \frac{ 2u\sqrt{ u } }{ 3 }$

$\frac{ 1 }{ 18 } \times \frac{ 2\left( 9{t}^{2}+4 \right)\sqrt{ 9{t}^{2}+4 } }{ 3 }$

$\frac{ \left( 9{t}^{2}+4 \right)\sqrt{ 9{t}^{2}+4 } }{ 27 }$

$\begin{array} { l }\frac{ \left( 9{t}^{2}+4 \right)\sqrt{ 9{t}^{2}+4 } }{ 27 }+C,& C \in ℝ\end{array}$

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