Evaluate $\int\frac{\sqrt{1+x^2}}{x}\, dx$ Evaluate $$\int\frac{\sqrt{1+x^2}}{x}\, dx$$
I tried substitution but failed miserably.
 A: Rewrite the integral as
$$\int \frac{x\sqrt{1+x^2}}{x^2}\,dx,$$ and let $u^2=1+x^2$.  Then $x\,dx=u\,du$, and we end up with
$$\int \frac{u^2}{u^2-1}\,du,$$
 which is probably familiar. Start by noting that $\frac{u^2}{u^2-1}=1+\frac{1}{u^2-1}$. 
A: Using Euler substitution by setting $\sqrt{1+x^2}=t-x$ then we have
$$
x=\frac{t^2-1}{2t}\qquad\Rightarrow\quad dx=\frac{1+t^2}{2t^2}\ dt.
$$
Hence
\begin{align}
\int\frac{\sqrt{1+x^2}}{x}\ dx&=\int\frac{\left(1+t^2\right)^2}{2t^2}\ dt\\
&=\int\left[\frac1{2}-\frac1{t+1}+\frac1{t-1}-\frac1{2t^2}\right]\ dt.
\end{align}
The rest part should be easy.
A: If we set $x=\sinh t$ we have:
$$\begin{eqnarray*} I &=& \int\frac{\cosh^2 t}{\sinh t}\,dt = \cosh t+\int\frac{dt}{\sinh t}=\cosh t+\int \frac{\cosh\frac{t}{2}}{\sinh\frac{t}{2}}\cdot\frac{dt}{2\cosh^2\frac{t}{2}}\\&=&\sqrt{1+x^2}+\log\tanh\frac{t}{2}=\sqrt{1+x^2}+\log\left(\frac{x}{1+\sqrt{x^2+1}}\right).\end{eqnarray*}$$
A: Try drawing a right triangle. Label one leg as "1" and the other leg as "x". We can solve for the hypotenuse of this triangle and get $\sqrt{1+x^{2}}$. Now assign a $\theta$ somewhere in the triangle so long as it is not the $90$ degree angle of our right triangle (I put my $\theta$ across from the leg marked "x"). If you do this, you should get the following relations: $$\sec(\theta)=\sqrt{1+x^{2}},\ \tan(\theta)=x,\ dx=\sec^{2}(\theta)d\theta$$ Now we can rewrite the integral as $$\int \! \frac{\sqrt{1+x^{2}}}{x} \, \mathrm{d}x=\int \! \frac{\sec(\theta)}{\tan(\theta)}sec^{2}(\theta) \, \mathrm{d}\theta=\int \! \sec^{2}(\theta)\csc(\theta) \, \mathrm{d}\theta$$ See if you can integrate it from here. It looks like one of those integrals that may need to be looked up in the back of a calc book
A: Let $x=\tan t$ so we have $$\int\frac{\sqrt{1+x^2}}{x} dx=\int\csc t dt+\int \frac{\sin t}{\cos^2 t}  dt=\ln|\csc t-\cot(t)|+\frac{1}{\cos t}+c$$
A: Put $$x = \tan y\, \Longrightarrow\,  dx = \sec^{2}y\,dy$$
Substituting this you will get $$I = \int \frac{sec^{3}y}{tany} dy$$
Now substitute $$\sec y = v\, \Longrightarrow\,  \sec y\tan y\,dy = dv\,  \Longrightarrow\, dy = \frac{dv}{\sec y\tan y}$$
Now $$I = \int \frac{v^3}{v(v^2-1)}dv$$
$$I = \int \frac{v^2}{v^2 - 1} dv$$
$$I = \int \frac{v^2-1 + 1}{v^2-1} dv$$
$$I = \int dv +\int \frac{1}{2} \left[\frac{1}{v-1} - \frac{1}{v+1}\right]dv$$
$$I = v + \frac{1}{2}\bigg[ \ln|v-1| - \ln|v+1|\bigg]+C.$$
