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There is an exercise on indefinite integral in some infinitesimal calculus book:

$$ \int \sqrt{x^{2} +1} \cdot dx $$

The solution uses the first substitution x = sinh u and after some transformations the book gets the result:

$$ \frac{1}{2} \cdot \left( x\sqrt{x^{2} +1} +\ln\left( x+\sqrt{x^{2} +1}\right)\right) +C $$

MAXIMA shows me a result:

$$ \[\frac{\operatorname{asinh}(x)}{2}+\frac{x\ \sqrt{{{x}^{2}}+1}}{2}\] $$

It seems to be same, but the question is: the book says that there is an another solution via substitution with tan. I tried to solve it with the tan-substitution and got very different result. Please, show me where is my error:

$$ \int \sqrt{x^{2} +1} \cdot dx=\int \sqrt{\tan^{2} a +1} \cdot \frac{da}{\cos^{2} a} =\int \sqrt{\frac{1}{\cos^{2} a}} \cdot \frac{da}{\cos^{2} a} =\int \frac{da\cdot \cos a}{\cos^{4} a} =\int \frac{d(\sin a)}{\left(\cos^{2} a\right)^{2}} = $$

$$ = \int \frac{d(\sin a)}{\left( 1\ -\ \sin^{2} a\right)^{2}} =\int \frac{dt}{\left( 1-t^{2}\right)^{2}} =\int \frac{e^{u} \cdot du}{\left( 1-e^{2\cdot u}\right)^{2}} =\frac{1}{2} \cdot \int \frac{2\cdot e^{u} \cdot du}{\left( 1-e^{2\cdot u}\right)^{2}} =\frac{1}{2} \cdot \int \frac{d\left( e^{2\cdot u}\right)}{\left( 1-e^{2\cdot u}\right)^{2}} = $$

$$ = \frac{1}{2} \cdot \int \frac{dz}{( 1-z)^{2}} =\frac{1}{2} \cdot \int \frac{d( z-1)}{( z-1)^{2}} =\frac{1}{2} \cdot \int \frac{dv}{v^{2}} =-\frac{1}{2\cdot v} +C $$

And then I'm trying to return back to the x through v -> z -> u -> t -> a -> x variables "back"-substitutions:

$$ = -\frac{1}{2\cdot ( z-1)} +C=-\frac{1}{2\cdot \left( e^{2\cdot u} -1\right)} +C=\frac{1}{2\cdot \left( 1-e^{2\cdot \ln t}\right)} +C=\frac{1}{2\cdot \left( 1-t^{2}\right)} +C=\frac{1}{2\cdot \left( 1-\sin^{2} a\right)} = $$

$$ = \frac{1}{2\cdot \cos^{2} a} +C=\frac{1}{2\cdot \cos(\arctan x) \cdot \cos(\arctan x)} +C=\frac{1}{2\cdot \frac{1}{\sqrt{1+x^{2}}} \cdot \frac{1}{\sqrt{1+x^{2}}}} +C=\frac{1+x^{2}}{2} +C $$

RandomB
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    The book&MAXIMA results are identical. Note that (up to an integration constant) $$\int\frac{1}{(1-t^2)^2},\mathrm dt=\frac{1}{4} \left(-\frac{2 t}{t^2-1}-\log (1-t)+\log (t+1)\right).$$ That might help you check where you went wrong. – Maximilian Janisch Aug 22 '20 at 15:16
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    Sorry, but it's impossible to follow your calculation if you don't put anything in the bound of the integral : $$\int \sqrt{x^2+1}\mathrm d x=\int\sqrt{\tan^2(x)+1}\cdot \frac{1}{\cos^2(x)},\mathrm d x,$$ has just no sense as written. It should be at least $$\int^y \sqrt{x^2+1}\mathrm d x=\int^{\arctan(y)}\sqrt{\tan^2(x)+1}\cdot \frac{1}{\cos^2(x)},\mathrm d x,$$ – Surb Aug 22 '20 at 15:16
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    You can use also reduction for secant and it gives same result. – zkutch Aug 22 '20 at 15:25
  • @Surb It's an indefinite integral; the real objection, I think, is that the calculation is hard to follow if we have to guess what the substitutions are (which bounds would help with). – Misha Lavrov Aug 22 '20 at 15:25

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You made a mistake going from $u$ to $z$: if $z = e^{2u}$, then $dz = 2e^{2u}\,du$, where you take it to be $2e^u\,du$. In fact, if you look at just what happens when you go from $t$ to $z$, you have replaced $t^2$ by $z$, but also replaced $dt$ by $dz$, where it should be $2\sqrt z\,dz$.

You could have stopped at $t$: once you have $\int \frac{dt}{(1-t^2)^2}$, you can take the partial fraction decomposition, and finish with no more substitutions.

Misha Lavrov
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