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How to find a primitive of $\left(x +\sqrt{1 + x^2}\right)^n$?

I have started it by parts but it never could end in a good position.

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    $\begingroup$ Maybe a trigonometric/hyperbolic substitution could be used? $\endgroup$
    – Khallil
    Commented Jul 22, 2015 at 19:34
  • $\begingroup$ @Khallil Benyattou: Can you tell what, sir? $\endgroup$
    – user142971
    Commented Jul 22, 2015 at 19:35
  • $\begingroup$ Had you tried a change of the kind $x+\sqrt{1+x^2}=u$? (Im not sure any way if this can work). $\endgroup$
    – Masacroso
    Commented Jul 22, 2015 at 19:39
  • $\begingroup$ Haven't actually tried it, but that's what the $\sqrt{1+x^2}$ screamed out if integration by parts didn't work. I'll give it a shot. $\endgroup$
    – Khallil
    Commented Jul 22, 2015 at 19:39

4 Answers 4

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$\sqrt{1+x^2}$ looks like a trig transformation, but with the "wrong" sign. So we try with hyperbolic substitution:

$$x=\sinh t$$ $$dx = \cosh t \,dt$$

$$\int (\sinh t + \cosh t)^n \cosh t\,dt$$ $$=\int (e^t)^n (e^t + e^{-t})/2\, dt=\frac12 \int e^{t(n+1)}+e^{t(n-1)}\,dt$$ $$=\frac12\left(\frac{e^{t(n+1)}}{n+1}+\frac{e^{t(n-1)}}{n-1}\right)$$

Now you just need $e^t = \sinh t + \cosh t = x+\sqrt{1+x^2}$ to get back to $x$.

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$$\int(x+\sqrt{1+x^2})^ndx$$

Use Euler substitution $\color{gray}{x=\frac{u^2-1}{2u},\;dx=\bigg(1-\frac{u^2-1}{2u^2}\bigg)}$

$$\int\frac{1}{2}(u^2+1)u^{n-2}du$$

Expanding the integrand $\color{gray}{(u^2+1)u^{n-2}}$ gives $\color{gray}{u^{n-2}+u^n}$

$$\frac{1}{2}\int(u^{n-2}+u^n)du$$

$$=\frac{u^{n-1}}{2(n-1)}+\frac{u^{n+1}}{2(n+1)}+C$$

$$\boxed{\color{red}{\frac{\bigg(\sqrt{x^2+1}+x\bigg)^n\bigg(n\sqrt{x^2+1}-x\bigg)}{n^2-1}+C}}$$

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    $\begingroup$ Can you tell me more about Euler substitution? I'm new to this term. $\endgroup$
    – user142971
    Commented Jul 22, 2015 at 19:45
  • $\begingroup$ en.wikipedia.org/wiki/Euler_substitution $\endgroup$
    – 3SAT
    Commented Jul 22, 2015 at 19:51
  • $\begingroup$ You have a typo on the penultimate step $\endgroup$ Commented Jul 22, 2015 at 19:56
  • $\begingroup$ I will fix it right now $\endgroup$
    – 3SAT
    Commented Jul 22, 2015 at 20:02
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Try substituting $u=\sinh^{-1}x$

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$
    – Mankind
    Commented Jul 22, 2015 at 20:32
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    $\begingroup$ @HowDoMarh.This is only intended as a hint as to how to proceed $\endgroup$ Commented Jul 22, 2015 at 20:43
  • $\begingroup$ But with no indication about the reason why one could select this substitution, the utility of the answer is zero. $\endgroup$
    – Did
    Commented Jul 22, 2015 at 21:24
  • $\begingroup$ @Did: it becomes obvious as soon as you try it. In any event the same method has already been demonstrated by Orion $\endgroup$ Commented Jul 22, 2015 at 21:31
  • $\begingroup$ Actually there are methods to choose these changes of variable, and to be sure before hand which ones will work. These methods are the interesting part, not to be given their result. $\endgroup$
    – Did
    Commented Jul 22, 2015 at 23:43
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start with $x=\tan u$:

\begin{align} \int (x+\sqrt{+x^2})^ndx&=\int (\sec u + \tan u)^n\sec^2u du\\ \end{align}

Now using integration by parts with $dz=(\sec u + \tan u)^n\sec u du$ and $v=\sec u$ we obtain \begin{align} I&=\int (\sec u + \tan u)^n\sec u \sec u du\\ &=\frac 1n \sec u (\sec u + \tan u)^n-\frac 1n \int(\sec u+\tan u)^n \sec u \tan u du \end{align} again we choose $dz$ as before and $v=\tan u$ to obtain

\begin{align} I&=\frac 1n \sec u (\sec u + \tan u)^n-\frac {1}{n^2}(\sec u+\tan u)^n \tan u +\frac{1}{n^2}I \end{align} Therefore $$I=\frac{(n \sec u-\tan (u)) (\tan (u)+\sec (u))^n}{n^2-1}+C$$ substituting back to $x$ we obtain $$I=\frac{(n \sqrt{1+x^2}-x) (x+\sqrt{1+x^2})^n}{n^2-1}+C$$

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