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solve: $$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$ I tried: $$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$

$x=\sin t$ $$\int \frac{\mathrm dt}{\cos t\sin t-\sin^2t}$$ And I can not continue from here.

an4s
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Aligator
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3 Answers3

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You can proceed from where you are like this, e.g.

\begin{eqnarray} \mathcal I &=& \int\frac{dx}{x\sqrt{x^2-1}\left(\sqrt{x^2-1}+x\right)} =\\ &=&-\int\frac{\sqrt{x^2-1}-x}{x\sqrt{x^2-1}}dx=\\ &=&-\int\frac1xdx +\int\frac1{\sqrt{x^2-1}}dx=\\ &=&-\log|x|+\log\left(\sqrt{x^2-1}+x\right)+C \end{eqnarray}

EDIT For the second integral use $t=\cosh x$, recalling that $\sinh^2 x= \cosh^2-1$, and $\operatorname{arccosh} x=\log(\sqrt{x^2-1}+x)$.

EDIT 2 Alternatively, as in comment, $x = \sec t$, brings the second integral to $\int \sec t dt =\log(\tan t + \sec t) + C= \log(\sqrt{\sec^2t-1}+\sec t)+C\dots$

dfnu
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    You can simplify it further to $\log\left(\left|\dfrac{\sqrt{x^2-1}}{x}+1\right|\right) + C$. – an4s Nov 29 '19 at 21:17
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    Thanks, I have a question How you solved $\int \frac{1}{\sqrt{x^2-1}}$? did you use $x=sect$ another answer for this integral is -Arcsin(x) right? – Aligator Nov 29 '19 at 21:25
  • @hgs Another way: substitute $x = \sec t$ to get $\int\frac{\sec t\tan t + \sec^2t}{\tan t + \sec t}\mathrm dt$, and then substitute $r = \tan t + \sec t$. – an4s Nov 29 '19 at 21:30
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    @hgs i just used $t=\operatorname{cosh}x$. – dfnu Nov 29 '19 at 21:32
  • @hgs In fact I have used $x=\sec(t)$ :-) with another way. – Sebastiano Nov 29 '19 at 21:37
  • @hgs, No I don't think you can write it with $-\arcsin x$ – dfnu Nov 29 '19 at 21:43
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Hint: Put $x＝\dfrac{1}{\cos t}$, then $$\int \frac{\cos^3 t}{\sin t(\sin t+1)}\frac{\sin t}{\cos^2 t}dt＝\int \frac{d\sin t}{\sin t+1}$$

Yan Peng
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Hint: You can try with $x=\sec(t)$ with a trigonometric substitution. With this substitution you will obtain $$\int \frac{1}{\sec (t)+\tan (t)}dt=\int \frac{1}{\frac{1}{\cos (t)}+\frac{\sin(t)}{\cos (t)}}dt$$ After another substitution $t=1+\sin (u)$ and $u=\mathrm{arcsec}(x)$ you will try the solution of the integral.

Sebastiano
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