Johnsy's question about finding an indefinite integral

Prove It

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MHB Math Helper
Evaluate \displaystyle \begin{align*} \int{ \frac{1}{\sqrt{5x} \, \left( 1 + 5x \right) } \, \mathrm{d}x} \end{align*}
The big clue here is the square root in the denominator, because \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{x} \right) = \frac{1}{2\,\sqrt{x}} \end{align*}. So this suggests that you probably need to find a square root function to substitute. Rewrite your integrand as

\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{5x} \, \left( 1 + 5x \right) } \, \mathrm{d}x} &= \int{ \frac{1}{\sqrt{5x} \, \left[ 1 + \left( \sqrt{5x} \right) ^2 \right] } \, \mathrm{d}x} \\ &= \frac{2}{5} \int{ \frac{1}{1 + \left( \sqrt{5x} \right) ^2 } \, \left( \frac{5}{2\,\sqrt{5x} } \right) \, \mathrm{d}x} \end{align*}

and now apply the substitution \displaystyle \begin{align*} u = \sqrt{5x} \implies \mathrm{d}u = \frac{5}{2\,\sqrt{5x} } \, \mathrm{d}x \end{align*} and the integral becomes \displaystyle \begin{align*} \frac{2}{5} \int{ \frac{1}{1 + u^2} \, \mathrm{d}u} \end{align*} which you should now be able to work with.