Integral of $\frac{\sqrt{x^2+1}}{x^2}$ How to integrate: $$\int \frac{\sqrt{x^2+1}}{x^2}\text dx$$
I've typed this into wolframalpha as a result I received something with inverse hyperbolic sine. Is it possible to have it without this hyperbolic sine? If so, how to do that?
 A: Let $x=\tan\theta$
$$\int\dfrac{\sqrt{1+x^2}}{x^2}dx=\int\dfrac{\sec^3\theta}{\tan^2\theta}d\theta$$
$$=\int\dfrac{\cos\theta}{(1-\sin^2\theta)\sin^2\theta}d\theta$$
Set $\sin\theta=u$
A: Consider the identity:
$$\sinh^{-1}x=\ln(x+\sqrt{1+x^2})$$
This is useful, because:
$$\begin{align}
\int \frac{\sqrt{x^2+1}}{x^2} 
&= \sinh^{-1}x - \frac{\sqrt{x^2+1}}{x} + C \\
&= \boxed{\ln(x+\sqrt{1+x^2}) - \frac{\sqrt{x^2+1}}{x} + C}
\end{align}$$
A: Let $$ I= \displaystyle \int \frac{\sqrt{x^2+1}}{x^2}dx = \int \frac{\sqrt{x^2+1}}{x^2} \times \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}dx $$
$$\displaystyle I  = \int\frac{x^2+1}{x^2\sqrt{x^2+1}}dx = \int\frac{1}{\sqrt{x^2+1}}dx+\int \frac{1}{x^2\sqrt{x^2+1}}dx$$
Now in second Integral , Put $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get   $$\displaystyle I  = \ln\left|x+\sqrt{x^2+1}\right|+J$$
So Integral $J$ Convert into $$\displaystyle -\int \frac{t}{\sqrt{t^2+1}}dt\;,$$ Now Substitute $t^2+1=u^2\;,$ Then $tdt = udu$
So  Second Integral $J$ into $$\displaystyle -\int \frac{udu}{u} = -u=-\sqrt{t^2+1}=-\frac{\sqrt{x^2+1}}{x}$$
So $$\displaystyle I = \frac{\sqrt{x^2+1}}{x^2}dx= \ln\left|x+\sqrt{x^2+1}\right|-\frac{\sqrt{x^2+1}}{x}+\mathcal{C}$$
A: Let $x=\tan\theta, dx=\sec^2\theta d\theta$ to get
$\displaystyle\int\frac{\sqrt{x^2+1}}{x^2}dx=\int\frac{\sec\theta}{\tan^2\theta}\cdot\sec^2\theta d\theta=\int\frac{\sec\theta}{\tan^2\theta}\big(\tan^2\theta+1\big)d\theta=\int\big(\sec\theta+\csc\theta\cot\theta\big)d\theta$
$=\displaystyle\ln\big|\sec\theta+\tan\theta \big|-\csc\theta+C=\ln\big(\sqrt{1+x^2}+x\big)-\frac{\sqrt{1+x^2}}{x}+C$
A: $$
\begin{aligned}
\int \frac{\sqrt{x^2+1}}{x^2} d x 
= & -\int \sqrt{x^2+1}\, d\left(\frac{1}{x}\right) \\
= & -\frac{\sqrt{x^2+1}}{x}+\int \frac{1}{\sqrt{x^2+1}} d x \quad (\textrm{ IBP })\\
= & -\frac{\sqrt{x^2+1}}{x }+\sinh ^{-1} x+C \quad (\textrm{ via }x=\sinh \theta)
\end{aligned}
$$
