Given that $y=(3x+1)\sqrt{6x-1}$, express $\frac{\mathrm{d}y}{\mathrm{d}x}$ in the form \(\

Question

Given that $y=(3x+1)\sqrt{6x-1}$, express $\frac{\mathrm{d}y}{\mathrm{d}x}$ in the form $\frac{kx}{\sqrt{6x - 1}}$, where $k$ is a constant. Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{9x}{\sqrt{6x - 1}} \, \mathrm{d}x.$

Accepted Answer

$k=27$, and $\displaystyle\int\frac{9x}{\sqrt{6x-1}}\,\mathrm dx=\frac13(3x+1)\sqrt{6x-1}+C$.

Applications of Integration - Integration as an Inverse Process of Differentiation

Applications of Integration - Integration as an Inverse Process of Differentiation

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