The fastest solution for $\int \sqrt{x^2+1}\,dx$

Question

The integral $(1)$ is typically attacked using trigonometric substitution, more specifically, using Case II, which involves substituting $x=a\tan{\theta}$. This method leads to having to evaluate the integral of secant cubed, which can be solved either by integration by parts or using the reduction formula. Here we provide an alternative approach to solve the integral $(1)$ and the like.

Let's evaluate the integral

$$\int\sqrt{x^2+1}\,dx\tag{1}$$

We will make use of the following general transformation formula:

$$\int f\left(x, \sqrt{x^2+a^2}\right)\, dx = \int f\left(\frac{e^{\theta}-e^{-\theta}}{2}a, \frac{e^{\theta}+e^{-\theta}}{2}a\right) \frac{e^{\theta}+e^{-\theta}}{2}a\, d\theta\tag{2}$$

Where $\theta=\sinh^{-1}(\frac{x}{a})$ and $a>0$.

Solution. From $(2)$ follows that

$$\begin{aligned}\int\sqrt{x^2+1}\,dx&=\frac14\int \left(e^\theta+e^{-\theta}\right)^2\, d\theta\\&=\frac14\int \left(e^{2\theta}+e^{-2\theta}+2\right)\,d\theta\\&=\frac{e^{2\theta}-e^{-2\theta}}{8}+\frac{\theta}{2}+C\\&=\frac{e^{2\sinh^{-1}(x)}-e^{-2\sinh^{-1}(x)}}{8}+\frac{\sinh^{-1}(x)}{2}+C \end{aligned}$$

Compare this solution with the one provided in the Blackpenredpen and Integrals for you channels. The closed forms can be proven equivalent to the one given here easily. Now, I consider that the solutions provided through these YouTube channels are much more complicated than the general solution I am proposing here for this type of integrals. Do you know of a simpler solution than mine? I don't consider the substitution $x=\sinh(u)$ to be simpler than this, as suggested by the number of questions in this forum asking for solutions that don't use trigonometric or hyperbolic substitutions.

This solution is part of a much more general method that can be seen as an alternative to trigonometric substitution (although it does more than that). You can go to the section 'An alternative method to trigonometric substitution' on my blog and refer to Example 5 for further illustration. For other cases you also have

$$\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{i\alpha}, \frac{1-e^{i\alpha}}{1+e^{i\alpha}}\right)\,\frac{e^{-i\alpha}-e^{i\alpha}}{2i}a\,d\alpha\tag{3}$$

$$\int f\left(x, \sqrt{x^2-a^2}, \frac{\sqrt{x-a}}{\sqrt{x+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, \frac{e^{-i\alpha}-e^{i\alpha}}{2}a, \frac{1-e^{i\alpha}}{1+e^{i\alpha}}\right)\,\frac{e^{-i\alpha}-e^{i\alpha}}{2i}a\,d\alpha\tag{4}$$

Where $\alpha=\cos^{-1}\left(\frac{x}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x}{a}\right)$, $\gamma=\sec^{-1}\left(\frac{x}{a}\right)$ and $\frac{x}{a}\geq1$.

Apart from the Euler substitutions, do you know of any other general approaches that allow getting rid of trigonometric or hyperbolic substitutions (which can be quite bothersome for students) that you could share? Thanks in advance.

Answer 1

Another possible approach based on the integration by parts method:

\begin{align*} \int\sqrt{x^{2} + 1}\mathrm{d}x & = x\sqrt{x^{2} + 1} - \int\frac{x^{2}}{\sqrt{x^{2} + 1}}\mathrm{d}x\\\\ & = x\sqrt{x^{2} + 1} - \int\sqrt{x^{2} + 1}\mathrm{d}x + \int\frac{1}{\sqrt{x^{2} + 1}}\mathrm{d}x \end{align*} which implies that \begin{align*} \int\sqrt{x^{2} + 1}\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^{2} + 1} + \text{arcsinh}(x)\right) + C \end{align*}

Hopefully this helps!

Answer 2

$\frac {e^{2\sinh^{-1} x} - e^{-2\sinh^{-1} x}}{8}$ looks ugly but it has a nice simplification: $$\frac {e^{2\sinh^{-1} x} - e^{-2\sinh^{-1} x}}{8} = \frac {x\sqrt {x^2+1}}{2}$$

You might notice that if we took a step back,

$$\frac {e^{2\theta} + e^{-2\theta}}{8} = \frac {(e^{θ}−e^{−θ})(e^{\theta}+e^{-\theta})}{8}$$

And when you reverse the substitution, just directly put $\frac {e^{\theta}-e^{-\theta}}{2}=x$ and $\frac {e^{\theta}+e^{-\theta}}{2}=\sqrt {x^2+1}$.

This might have been more obvious if you worked it all the way through with hyperbolic functions:

$$\begin{align}\int \cosh^2 t\mathrm dt&= \int \frac 12 (\cosh 2t + 1)\mathrm dt\\&= \frac {\sinh 2t}{4} + \frac {t}{2} + C\\&= \frac {\cosh t\sinh t + t}{2} + C(\because \sinh 2t = 2\cosh t\sinh t)\end{align}$$

Answer 3

I don’t think this is the fastest method, but I still thought of sharing an alternative perspective. The integrand can be rewritten as $x^0(x^2+1)^\frac12$ which is of the form of Chebyshev’s differential binomial. Hence, substituting $t=\sqrt{1+\frac1{x^2}}$,

$$\mathrm dx=-x^3t\mathrm dt, \sqrt{x^2+1}=xt\text{sgn}x$$ $$\begin{align}\implies \int\sqrt{x^2+1}\mathrm dx&=-\text{sgn}x\int t^2x^4\mathrm dt\\&= -\text{sgn}x\int\frac{t^2}{t^4-2t^2+1}\mathrm dt\end{align}$$

This can now be finished by writing $t^2=\frac12(t^2+1)+\frac12(t^2-1)$.

Answer 4

Performing the hyperbolic substitution $x=\sinh t$:

$$\begin{align}\int\sqrt{x^2+1}\mathrm dx&=\int(x^2+1)\frac{\mathrm dx}{\sqrt{x^2+1}}\\&=\int\cosh^2t\mathrm dt\\&=\frac12\int1+\cosh2t\mathrm dt\\&=\frac{t}2+\frac{\sinh t\cosh t}2+C\\&=\frac{\sinh^{-1}(x)+x\sqrt{x^2+1}}2+C\end{align}$$

Answer 5

Substitute $t=x+\sqrt{x^2+1}\implies\dfrac{\mathrm dt}t=\dfrac{\mathrm dx}{\sqrt{x^2+1}}, \dfrac{t+\frac1t}2=\sqrt{x^2+1}, \dfrac{t-\frac1t}2=x:$

$$\begin{align}\int\sqrt{x^2+1}\,\mathrm dx&=\frac14\int\left(t+\frac1t\right)^2\frac{\mathrm dt}t\\&=\frac14\int t+\frac2t+\frac1{t^3}\,\mathrm dt\\&=\frac18\left(t^2-\frac1{t^2}\right)+\frac12\ln t+C\\&=\frac{\left(\dfrac{t+\frac1t}2\right)\left(\dfrac{t-\frac1t}2\right)}2+\frac12\ln t+C\\&=\frac{x}2\sqrt{x^2+1}+\frac12\sinh^{-1}x+C\end{align}$$