I tried the Euler substitution: $t=x+\sqrt{x^{2}+a}$ and I got $\int \frac{4ta}{(t^{2}+a)^{2}}dt$ and I don't know wnat now or if it is the right way.
Thank you in advance.
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2Now use substitution $t^2+a=u$, $2tdt=du$ – Vasili Dec 13 '17 at 18:40
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1Try $x=a\tan y$ – lab bhattacharjee Dec 13 '17 at 18:41
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1Substitute $$x=a\tan(t)$$ – Dr. Sonnhard Graubner Dec 13 '17 at 18:41
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Or $x=a\sinh y$. Or integrate by parts. – Bernard Dec 13 '17 at 18:52
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substituting $$x=a\tan(t)$$ we get $$a^2+x^2=a^2(1+\tan^2(t))$$ and $$dx=a(1+\tan^2(t))dt$$ thus our integral is $$\int\frac{a^2(1+\tan^2(t))}{(a^2(1+\tan^2(t))^{3/2}}dt$$ this gives $$|a|\int\sqrt{\frac{sin^2(t)+\cos^2(t)}{\cos^2(t)}}dt=\int\pm\frac{a}{\cos(t)}dt$$ note that $$\frac{1}{\cos^2(t)}=\frac{\sin^2(t)+\cos^2(t)}{\sin^2(t)}=1+\tan^2(t)$$
Dr. Sonnhard Graubner
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Could you give some more details? How did you get the numerator below "our integral is..."? – Karol Borkowski Dec 13 '17 at 19:19
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