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

Question

So I've seen some options on the internet that are fairly good, but I have this substitution: $x^2+1=t-x$, you square both sides and get $x = (t^2-1)/t$ and $x + 1 = (t^2-1)/2t + 1$. If we call that function $e(t)$ for example, $dx$ is equal to $e(t)$ derivative. How to proceed from here?

Answer 1

this is one of the "famous" integrals. It is written as :

$ \int \frac {1}{\sqrt{x^2 +1}} = sinh^{-1}(x) + c $

Where : $ arcsinh(x) = ln(x + \sqrt{x^2 + 1}) $ and $ sinh$ is called the hyperbolic sine.

Let me know if you want me to post a complete solution by scratch, without supposing you know hyperbolic functions.

Answer 2

The appearance of the expression $\sqrt{1 + x^2}$ suggests the standard trigonometric substitution $$\boxed{x = \tan \theta, \qquad dx = \sec^2 \theta \,d\theta} ,$$ which leads to integrating $\int \sec \theta \,d\theta$.

On the other hand, if you're familiar with hyperbolic functions, it also suggests $$\boxed{x = \sinh t, \qquad dx = \cosh t \,dt} ,$$ which leads almost immediately to the antiderivative.

Answer 3

Analytic method :

Substitute : $x=tan(u) $ and $dx=sec^2(u)du$. Then $ \sqrt{x^2+1} = \sqrt{tan^2(u) + 1}=sec(u) $ and $ u=arctan(x) $

Then you get : $ \int sec(u)du $

Multiply numerator and denominator of $ sec(u) $ by $ tan(u) + sec(u) $ and you get :

$ \int \frac{sec^2(u) + tan(u)sec(u)}{tan(u) + sec(u)} du $

For the integrand , substitute $ s=tan(u) + sec(u) $ and $ ds = (sec^2(u) + tan(u)sec(u))du$

$ \int 1/s ds = ln(s) + c = ln(tan(u) + sec(u)) + c = ln(tan(arctan(x) + sec(arctan(x)) = ln(x + \sqrt{x^2 +1}) + c = arcsinh(x) + c $

Answer 4

Substituting $x=\frac{t^2-1}{2t}$, we get $dx= \frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$

Plugging this in our integral gives: $$\int \frac{\frac{1}{2}(1+\frac{1}{t^2})dt}{\sqrt{1+\left(\frac{t^2-1}{2t}\right)^2}}$$ $$\text {(Skipped some algebra)}$$ $$=\int \frac{(t^2+1)dt}{t^2\sqrt{\left(\frac{t}{2}+\frac{1}{2t}\right)^2}}$$ $$\text {(Skipped some algebra) }$$ $$=\int \frac{dt}{t}$$ $$=\ln t +C$$ Although I recommend that instead of doing this and creating a mess, you substitute $x=\tan \theta$ in the original integral (it's much cleaner than this.)