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I was wondering if the solutions in the end are equal for this specific integral : $\int \frac{1}{\sqrt{1+4x-5x^2}}\,dx$

Since this is the integral of the form $\int R(x,\sqrt{ax^2+bx+c})$ we can use Euler substitutions. In this case $a<0$ but $c>0$ so we can use $\sqrt{1+4x-5x^2}=1+tx$. With this type of substitution I'm getting a new integral of rational functions expressed through $t$:

$-\int \frac{-2t^2+8t+6}{(t^2+5)(t-5)(t+1)}\,dx$

The solution is expressed through logarithms and arctan.

The solution at the back of my book and the solution that is used by most online calculators is the following :

$\frac{1}{\sqrt{5}}\arcsin(\frac{5x}{3}-\frac{2}{3})+C$

This is done by completing the square under the root and getting an integral of the form $\int \sqrt{a^2-x^2}\, dx$

So are the two solutions legitimate?

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    $\begingroup$ Can't you just differentiate both and see if you arrive at the original expression? $\endgroup$ – Alex Provost Mar 1 '18 at 12:13
  • $\begingroup$ That's a bit harder than it looks. Even with wolfram. The first solution is just a mess. Was looking more for an answer whether both solutions are reasonable and if everything is correct then they should be equal in the end. $\endgroup$ – DreaDk Mar 1 '18 at 12:24
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You have made an error using the Euler substitution. It should have ended up with: $$ -\int\frac{2dt}{t^2+5}=-\frac{2}{\sqrt5}\arctan\frac{t}{\sqrt5}. $$ Upon substitution: $$ t=\frac{\sqrt{1+4x-5x^2}-1}{x} $$ it gives (up to a constant) the same expression as $\frac{1}{\sqrt{5}}\arcsin(\frac{5x}{3}-\frac{2}{3})$.

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