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# integral of $9*t^4$/ √$9t^4 + 4t^2$

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## Answer to a math question integral of $9*t^4$/ √$9t^4 + 4t^2$

Darrell
4.5
$9 \times \int{ \frac{ {t}^{4} }{ \sqrt{ 9{t}^{4}+4{t}^{2} } } } \mathrm{d} t$
$9 \times \int{ \frac{ {t}^{4} }{ t\sqrt{ 9{t}^{2}+4 } } } \mathrm{d} t$
$9 \times \int{ \frac{ {t}^{3} }{ \sqrt{ 9{t}^{2}+4 } } } \mathrm{d} t$
$9 \times \int{ \frac{ u-4 }{ 162\sqrt{ u } } } \mathrm{d} u$
$9 \times \frac{ 1 }{ 162 } \times \int{ \frac{ u-4 }{ \sqrt{ u } } } \mathrm{d} u$
$\frac{ 1 }{ 18 } \times \int{ \frac{ u-4 }{ \sqrt{ u } } } \mathrm{d} u$
$\frac{ 1 }{ 18 } \times \int{ \frac{ u-4 }{ {u}^{\frac{ 1 }{ 2 }} } } \mathrm{d} u$
$\frac{ 1 }{ 18 } \times \int{ \frac{ u }{ {u}^{\frac{ 1 }{ 2 }} }-\frac{ 4 }{ {u}^{\frac{ 1 }{ 2 }} } } \mathrm{d} u$
$\frac{ 1 }{ 18 } \times \int{ {u}^{\frac{ 1 }{ 2 }}-\frac{ 4 }{ {u}^{\frac{ 1 }{ 2 }} } } \mathrm{d} u$
$\frac{ 1 }{ 18 } \times \left$\int{ {u}^{\frac{ 1 }{ 2 }} } \mathrm{d} u-\int{ \frac{ 4 }{ {u}^{\frac{ 1 }{ 2 }} } } \mathrm{d} u \right$$
$\frac{ 1 }{ 18 } \times \left$\frac{ 2u\sqrt{ u } }{ 3 }-\int{ \frac{ 4 }{ {u}^{\frac{ 1 }{ 2 }} } } \mathrm{d} u \right$$
$\frac{ 1 }{ 18 } \times \left$\frac{ 2u\sqrt{ u } }{ 3 }-8\sqrt{ u } \right$$
$\frac{ 1 }{ 18 } \times \left$\frac{ 2\left( 9{t}^{2}+4 \right$\sqrt{ 9{t}^{2}+4 } }{ 3 }-8\sqrt{ 9{t}^{2}+4 } \right)$
$\frac{ \left$9{t}^{2}+4 \right$\sqrt{ 9{t}^{2}+4 } }{ 27 }-\frac{ 4 }{ 9 } \times \sqrt{ 9{t}^{2}+4 }$
$\begin{array} { l }\frac{ \left$9{t}^{2}+4 \right$\sqrt{ 9{t}^{2}+4 } }{ 27 }-\frac{ 4 }{ 9 } \times \sqrt{ 9{t}^{2}+4 }+C,& C \in ℝ\end{array}$

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