Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?

The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.

And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?

share|cite|improve this question
WA now seems to understand LaTeX and so it is easy to experiment and see a general form. Try… and change the exponent. – lhf Sep 14 '11 at 11:39
Where did $a$ come from? – TonyK Sep 14 '11 at 11:46
There is a nice recursion you can derive: letting $\mu_n=\int \frac{t^n}{\sqrt{t^2+1}}\mathrm dt$, we have $$\mu_n=\frac1{n}(t^{n-1}\sqrt{1+t^2}-(n-1)\mu_{n-2})$$. The integrals for $n=0,1$ are easily derived, so you can use those to start the recursion. – J. M. Sep 14 '11 at 11:53
but how did you get this recursion? – Charles Bao Sep 14 '11 at 12:11

Recall the hyperbolic functions $$\cosh t= \frac{e^t + e^{-t}}{2} = \cos(it)$$ and $$\sinh t=\frac{e^t - e^{-t}}{2} = i\sin(-it).$$

Note that $\frac{d}{dt}\sinh t = \cosh t$, $\frac{d}{dt}\cosh t = \sinh t$ and also $\cosh^2 t -\sinh^2 t = 1$.

Making the substitution $\sinh t=x $ we see that
$$\frac{x^n\, dx}{\sqrt{1+x^2}} = \frac{\sinh^n t\, \cosh t\,dt}{\sqrt{1+\sinh^2t}}= \frac{\sinh^n t\, \cosh t\,dt}{\sqrt{\cosh^2t}}=\sinh^n t\, dt$$ which leads us to $$\int\frac{x^n\, dx}{\sqrt{1+x^2}} = \int \sinh^n t\, dt.$$ To complete the problem, the binomial theorem is useful.

share|cite|improve this answer
This is one case where the hyperbolic sine is a better substitution choice than the tangent. – J. M. Sep 14 '11 at 12:24
@J.M.: It is fairly standard in the hyperbolic world :) – AD. Sep 14 '11 at 12:26
@J. M.: The integration is definitely easier with $x=\sinh(t)$, but it often leaves a more involved back-substitution to get things in terms of $x$. – robjohn Sep 14 '11 at 13:34
@robjohn: Dunno, but I don't find $\mathrm{arsinh}(x)$ complicated... – J. M. Sep 14 '11 at 13:36
@robjohn: That is why we get the $\log$ etc. – AD. Sep 14 '11 at 13:38

Here's one way to go about deriving a recursion relation for integrals of the form

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

Split the integral like so:

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

and integrate by parts:

$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\sqrt{1+x^2} x^{n-2}\mathrm dx$$

Noting that $1+x^2$ is always positive for real $x$, we then complicate things a little:

$$\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int(1+x^2)\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$

Perform another split:

$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\left(\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx+\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$

and we see something we can isolate:

$$n\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$

and then we finally divide both sides by $n$:

$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=\frac1{n}\left(x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$

We can use the starting values $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\mathrm{arsinh}\,x$ and $\int\frac{x \mathrm dx}{\sqrt{1+x^2}}=\sqrt{1+x^2}$ for the recursion.

(This is a response to Srivatsan's comment, which got too long for the comment box.)

share|cite|improve this answer
Thanks for writing the answer. (Btw I ran out of votes, so I'll upvote it when I get them back.) – Srivatsan Sep 14 '11 at 15:12
@Sri: No worries; I expend my votes too quickly, too. – J. M. Sep 14 '11 at 15:19

I would first try the substitution $x=\tan(\theta)$, so that $\sqrt{1+x^2}=\sec(\theta)$. That gives $$ \begin{align} \int\frac{x^n}{\sqrt{1+x^2}}\;\mathrm{d}x &=\int \tan^n(\theta)\sec(\theta)\;\mathrm{d}\theta\\ &=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int\tan^{n-2}(\theta)\;\sec^3(\theta)\;\mathrm{d}\theta\\ &=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int(\tan^n(\theta)+\tan^{n-2}(\theta))\;\sec(\theta)\;\mathrm{d}\theta\\ &=\frac{1}{n}\tan^{n-1}(\theta)\sec(\theta)-\frac{n-1}{n}\int\tan^{n-2}(\theta)\;\sec(\theta)\;\mathrm{d}\theta \end{align} $$ If $n$ is odd, this reduces to $$ \int\tan(\theta)\sec(\theta)\;\mathrm{d}\theta=\sec(\theta)+C $$ If $n$ is even, this reduces to $$ \begin{align} \int\sec(\theta)\;\mathrm{d}\theta&=\int\sec^2(\theta)\;\mathrm{d}\sin(\theta)\\ &=\int\frac{1}{2}\left(\frac{1}{1-\sin(\theta)}+\frac{1}{1+\sin(\theta)}\right)\;\mathrm{d}\sin(\theta)\\ &=\frac{1}{2}\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)+C\\ &=\log(\sec(\theta)+\tan(\theta))+C \end{align} $$

share|cite|improve this answer

Since $\frac{d}{dt}\sqrt{1+t^2} = \frac{t}{\sqrt{1+t^2}}$, we can integrate by parts to get $$ \int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt = \int t\cdot \frac{t}{\sqrt{1+t^2}}\mathrm dt = t\sqrt{1+t^2} - \int \sqrt{1+t^2}\mathrm dt. $$ Cheating a little bit by looking at a table of integrals, we get that since $$ \frac{d}{dt} \left [ t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2}) \right ] = t\frac{t}{\sqrt{1+t^2}} + \sqrt{1+t^2} + \frac{1}{t + \sqrt{1+t^2}} \left [ 1 + \frac{t}{\sqrt{1+t^2}} \right ] $$ which simplifies to $2\sqrt{1+t^2}$, the integral on the right above is $\frac{1}{2}[t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2})]$ and thus we have $$ \int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt = \frac{1}{2}\left [t\sqrt{1+t^2} - \ln(t + \sqrt{1+t^2})\right ] $$ which matches the answer given by Charles Bao if we set $X=x$ and $a=1$ in his original post.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.