How can I prove that the indefinite integral
$ \int\frac{\sqrt{a+bu}}{u}du $
is equal to
$ 2\sqrt{a+bu} +a \int\frac{1}{u\sqrt{a+bu}}du$
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1Take the derivative. If you can show it's equivalent to the original integrand, then you have proved they are equivalent. – Jared Jul 18 '14 at 03:47
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good point they are equal. – Rene Schipperus Jul 18 '14 at 03:50
1 Answers
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$\int\frac{\sqrt{a+bu}}{u}du=\int\frac{a+bu}{u\sqrt{a+bu}}du=\int\frac{b}{\sqrt{a+bu}}du+\int\frac{a}{u\sqrt{a+bu}}du=2\sqrt{a+bu}+\int\frac{a}{u\sqrt{a+bu}}du$
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