How do you integrate #int (4x)/sqrt(3x)# using substitution?
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However, we should see that substitution is a waste of time:
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To integrate ( \int \frac{4x}{\sqrt{3x}} ) using substitution, let's set ( u = \sqrt{3x} ). Then, we can find ( du ) in terms of ( dx ) and rewrite the integral in terms of ( u ). Here's how:

Let ( u = \sqrt{3x} ).

Find ( du ): [ du = \frac{d}{dx}(\sqrt{3x})dx ] [ du = \frac{1}{2\sqrt{3x}} \cdot 3dx ] [ du = \frac{3}{2\sqrt{3x}}dx ]

Rewrite the integral in terms of ( u ): [ \int \frac{4x}{\sqrt{3x}}dx = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{\sqrt{x}} dx ] [ = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{\sqrt{3x}} \cdot \sqrt{3} dx ] [ = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{u} \cdot \sqrt{3} du ] [ = 4\int \frac{1}{u} du ] [ = 4 \lnu + C ]

Substitute back for ( u ): [ = 4 \ln\sqrt{3x} + C ] [ = 4 \ln(\sqrt{3})(\sqrt{x}) + C ] [ = 4 \ln(\sqrt{3}) + 4 \ln(\sqrt{x}) + C ] [ = 2 \ln(3) + 2 \ln(x) + C ]
So, the integral of ( \frac{4x}{\sqrt{3x}} ) using substitution is ( 2 \ln(3) + 2 \ln(x) + C ), where ( C ) is the constant of integration.
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