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∫ √9x + 1 dx

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Answer to a math question ∫ √9x + 1 dx

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Madelyn
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$\int{ \frac{ 1 }{ 9 } \times \sqrt{ t } } \mathrm{d} t$
$\frac{ 1 }{ 9 } \times \int{ \sqrt{ t } } \mathrm{d} t$
$\frac{ 1 }{ 9 } \times \int{ {t}^{\frac{ 1 }{ 2 }} } \mathrm{d} t$
$\frac{ 1 }{ 9 } \times \frac{ 2t\sqrt{ t } }{ 3 }$
$\frac{ 1 }{ 9 } \times \frac{ 2\left( 9x+1 \right)\sqrt{ 9x+1 } }{ 3 }$
$\frac{ 2\sqrt{ 9x+1 }\left( 9x+1 \right) }{ 27 }$
$\begin{array} { l }\frac{ 2\sqrt{ 9x+1 }\left( 9x+1 \right) }{ 27 }+C,& C \in ℝ\end{array}$

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x(squared) -8x=0