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Use substitution to find the indefinite integral
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\[\int\limits_{?}^{?} \sqrt{5x+1} dx\]
Let \(u = 5x+1, ~du = 5dx \Rightarrow \dfrac{du}{5} = dx\)
\[\cdots~~~=\dfrac{1}{5}\int\sqrt{u}~du\]
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Yes, thank you
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Glad I helped.
So how would I start \[\int\limits 2\cos(2x) dx\]
Let \(\displaystyle u = 2x, ~du = 2dx \Rightarrow\int2\cos(2x)dx = \int\cos(u)du\)
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