Evaluating $\int \frac{1}{\sqrt{x^2 + a^2}}\, dx$ without resorting to trigonometric $u$-substitution I am looking for a quick and intuitive way to evaluate this indefinite integral without resorting to any trigonometric functions. I'm not sure if it is at all possible to do so, but I was just wondering if the integration can be done in a more straightforward manner. This integral arises in the problem of calculating the electric potential of a finite line charge.
$$\int \dfrac{1}{\sqrt{x^2 + a^2}}\, dx$$
 A: The hyperbola $H: 1+x^2=y^2$ can be parametrized using rational functions. More precisely, we have the point $(0,1)$ on it. The family of lines passing through $(0,1)$ is $x=t(y-1)$. Each of those lines intersects $H$ in two points, one of which is $(0,1)$. To find the other point, we substitute $x=t(y-1)$ in the equation defining $H$:
$$1+t^2(y-1)^2 = y^2$$
This is a second degree polynomial in $y$, whose roots are the $y$-coordinates of the intersection points of $Y$ with the line $x=t(y-1)$. One of those roots is $y=1$, so the constant term (once we have divided out by the leading coefficient), which is the product of the roots, must give the $y$-coordinate of the other point, call it $(x(t), y(t))$:
$$y(t) = \frac{t^2+1}{t^2-1}.$$
Since the point lies on $x=t(y-1)$, we have
$$x(t) = t\left(\frac{t^2+1}{t^2-1}-1\right) = \frac{2t}{t^2-1}.$$
Thus we have discovered the formula
$$1 +\left(\frac{2t}{t^2-1}\right)^2 = \left(\frac{t^2+1}{t^2-1}\right)^2.$$
In your integral (assume $a=1$ WLOG), we make the substitution $x = x(t)$. Then $$dx = \frac{2(t^2-1)-4t^2}{(t^2-1)^2} dt = -2\frac{1+t^2}{(t^2-1)^2}dt.$$
Thus
$$\int \frac{dx}{\sqrt{1+x^2}} = -2 \int\left(\frac{t^2-1}{t^2+1}\right)\frac{1+t^2}{(t^2-1)^2}dt = -2 \int\frac{dt}{1-t^2} = \int\left(\frac{1}{t-1} + \frac{1}{t+1}\right) dt$$ $$ = \log(t-1) + \log(t+1)$$
Now it seems like there are no trig functions involved, but really if you want to get $t$ in terms of $x$ you have to solve the quadratic $(t^2-1)x = 2t$, and you will actually end up with non-positive values inside the log. Using the complex log will make multiples of $\pi$ appear, as one should expect from a trigonometric integral...
Of course the trigonometric substitutions do essentially the same thing, because the hyperbola $1+x^2=y^2$ is also parametrized by sec and csc.
A: We can cheat and use a "magic" substitution. Let $t=x+\sqrt{a^2+x^2}$. Then $dt=\left(1+\frac{x}{\sqrt{a^2+x^2}}\right)\,dx=\frac{t}{\sqrt{a^2+x^2}}\,dx$.
Thus our integral is 
$$\int \frac{dt}{t}.$$
A: I don't know if this is good enough for you, but you don't have to do a substitution if you simply remember that the derivative of $\sinh^{-1} x$ is $\frac1{\sqrt{x^2+1}}$.
A: You could try $x=a(e^u - e^{-u})/2$ but this isn't going to give you the integral as an explicit function of $x$, and of course it's a long-winded way of substituting $x=a\sinh(u)$. The latter does seem like the simplest way, and gives you an explicit function as the answer. But here's the long form
$$x=a(e^u - e^{-u})/2 \implies dx=a(e^u + e^{-u})/2\;du$$
$$\begin{align}
\int\frac{1}{\sqrt{x^2+a^2}} dx &= \int\frac{1}{\sqrt{(a(e^u - e^{-u})/2)^2+a^2}} \frac{a}{2}(e^u + e^{-u})du \\
&= \int \frac{e^u + e^{-u}}{\sqrt{(e^{2u} -2 +e^{-2u}) +4}}du\\
&= \int \frac{e^u + e^{-u}}{\sqrt{e^{2u} +2 +e^{-2u}}}du\\
&= \int \frac{e^u + e^{-u}}{\sqrt{(e^u +e^{-u})^2}}du\\
&= \int du\\
&= u + C
\end{align}$$
You could argue this isn't using trigonometric functions, although it's equivalent to doing so, with the disadvantage that you can't get the integral explicitly. To be honest I'd always been kind of upset at having to deal with $\sinh$ and $\cosh$ as well as $\sin$ and $\cos$ but having written the above I think I'm appreciating them some more.
With (hyperbolic) trigonometric functions it's just
$$x = a\sinh u \implies dx = a\cosh du $$
$$\begin{align}
\int\frac{1}{\sqrt{x^2+a^2}} dx &= \int \frac{1}{\sqrt{a^2\sinh^2 u +a^2}}a\cosh du\\
&= \int \frac{\cosh u}{\cosh u} du &\because \sinh^2 \theta + 1 = \cosh^2\theta\\
&= \int du\\
&= u + C \\
&= \sinh^{-1} x + C
\end{align}$$
A: Although the two standard substitutions are:


*

*the trigonometric substitution $x=a\tan t$

*the hyperbolic substitution $x=a\sinh t$,
since the integrand is a rational function of $x$ and $\sqrt{Ax^2+Bx+C}$ you can always use an algebraic substitution called the Euler substitution 
\begin{equation*}
\sqrt{x^{2}+a^{2}}=x+t\Leftrightarrow t=\sqrt{x^{2}+a^{2}}-x.
\end{equation*}
As a matter of fact, since
\begin{equation*}
x=\frac{a^{2}-t^{2}}{2t}
\end{equation*}
and 
\begin{equation*}
dx=-\frac{a^{2}+t^{2}}{2t^{2}}dt
\end{equation*}
we have that
\begin{eqnarray*}
\int \frac{dx}{\sqrt{x^{2}+a^{2}}} &=&\int \frac{1}{\frac{a^{2}-t^{2}}{2t}+t}
\left( -\frac{a^{2}+t^{2}}{2t^{2}}\right) dt=-\int \frac{1}{t}dt \\
&=&-\ln \left\vert t\right\vert +C=-\ln \left\vert \sqrt{x^{2}+a^{2}}
-x\right\vert +C.
\end{eqnarray*}
A: As far as I can see there are really only two possibilities if hyperbolic functions are excluded.  Substituting $x=a\tan\theta$ gives
$$\int\sec\theta\,d\theta\ ,$$
which is probably the hardest "easy" trig integral, and $x=a\cot\theta$ gives
$$-\int\csc\theta\,d\theta$$
which is essentially the same.  In neither case could the subsequent calculations be regarded as especially easy, so I would suggest that there really is no good answer to the OP's question - you just have to grit your teeth and do the trig substitution after all.
A: Here's a way to do it. Write
$$y =  \int_0^{x(y)} \frac{1}{(t^2+1)^{1/2}} dt$$
and take a derivative with respect to $y$ in order to get
$$1 = \frac{1}{(x(t)^2+1)^{1/2}} x'(y)$$
or
$$x'(y) = (x(y)^2+1)^{1/2}.$$
Take a derivative again and substitute in for what $x'(y)$ is:
$$x''(y) = \frac{x'(y)x(y)}{(x(y)^2+1)^{1/2}} = x(y).$$
Now note that the first expression gives $x(0)=0$ and the second gives $x'(0)=1$. Thus, using typical constant coefficient linear ODE theory, we get $x(y) = \frac{1}{2} (e^y-e^{-y})=\sinh(y)$. Hence $y=\sinh^{-1} (x)$.
Is this intuitive? Well... it can't hurt to see if $x(y)$ is somehow a nicer function.
A: $\Rightarrow \displaystyle \int \frac{1}{\sqrt{x^2+a^2}}dx\;,$ Let $x^2+a^2 = y^2\;,$ Then $\displaystyle 2xdx = 2ydy\Rightarrow \frac{dx}{y} = \frac{dy}{x}$
Now Using ration and Proportion, We Get $\displaystyle \frac{dx}{y} = \frac{dy}{x} = \frac{d(x+y)}{(x+y)}$
Now our Integral convert into  $\displaystyle \int\frac{dx}{y} = \frac{d(x+y)}{(x+y)} = \ln \left|x+y\right|+\mathbb{C}$
So $\displaystyle \int\frac{1}{\sqrt{x^2+a^2}}dx = \ln \left|x+\sqrt{x^2+a^2}\right|+\mathbb{C}$
