I need to integrate the following using substitution:

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

My textbook has a similar example:

$$ \int \sqrt{x^2+1}\;x^5\;dx $$

They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:

$$ \int \sqrt{x^2+1}\;x^4\cdot x\;dx\\ =\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\ =\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\ =\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\ =\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C $$

So far so good. But when I try this method on the given integral, I get the following:

$$ \int x^2\sqrt{x^2+1}\;dx\\ =\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\ =\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\ =\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du $$

Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.

How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.

  • 2
    $\begingroup$ $\sqrt{u(u-1)}=\frac12\sqrt{(2u-1)^2-1}$, then you need a trig substitution $\endgroup$
    – Empy2
    Aug 13, 2021 at 9:12
  • 8
    $\begingroup$ You could simply start with a trig substitution. $\sqrt{x^2+1}$ suggests $x=\tan(u)$. $\endgroup$
    – robjohn
    Aug 13, 2021 at 9:25
  • $\begingroup$ Or you could at first try $x^{2}=(u-1)/2$, then you'll get a familiar integral. $\endgroup$
    – Mourad
    Aug 13, 2021 at 9:35
  • 1
    $\begingroup$ In the first step of your textbooks solution, shouldn't there be a $1/2$? $\endgroup$ Aug 13, 2021 at 10:36
  • 1
    $\begingroup$ alternatively, one could make the hyperbolic substitution $x=\sinh(t)$ and then use the fact that $\sinh(2t)=2\sinh(t)\cosh(t)$ and $\sinh^2(t)=\frac{1}{2}(\cosh(2t)-1)$. $\endgroup$
    – C Squared
    Aug 13, 2021 at 12:17

4 Answers 4


One way to proceed is: \begin{align} \int x^2 \sqrt{x^2+1} \, dx &= \frac{1}{2} \int \sqrt{u} \sqrt{u-1} \, du \qquad (u = x^2 + 1) \\ &= - \frac{1}{16} \int \frac{y^8 - 2 y^4 + 1}{y^5} \, dy \qquad (y = \sqrt{u} - \sqrt{u-1}) \\ &= - \frac{y^4}{64} + \frac{1}{64 y^4} + \frac{\log(y)}{8} + C. \end{align}

Here, to change the integration variable from $u$ to $y$, one needs \begin{equation} \frac{dy}{du} = \frac{1}{2} \left( \frac{1}{\sqrt{u}} - \frac{1}{\sqrt{u-1}} \right) = - \frac{y}{2 \sqrt{u} \sqrt{u-1}}, \end{equation} and also the expression of $u$ in terms of $y$, which can be obtained by squaring the both sides of $\sqrt{u-1} = \sqrt{u}-y$ and then solving it for $\sqrt{u}$: \begin{equation} u = \frac{(y^2+1)^2}{4y^2}. \end{equation} So, after the variable transformation from $u$ to $y$, one gets a factor $u(u-1)$, which is rewritten as \begin{equation} u(u-1) = \frac{(y^2+1)^2(y^2-1)^2}{16 y^4}. \end{equation}

Now, after performing the integration, one needs the following substitutions \begin{equation} y = \sqrt{x^2 + 1} - x , \qquad y^4 - \frac{1}{y^4} = - 8 x (2 x^2 + 1) \sqrt{x^2 + 1}, \end{equation} to get the final result in $x$: \begin{equation} \int x^2 \sqrt{x^2+1} \, dx = \frac{1}{8} \left[ x (2x^2+1) \sqrt{x^2+1} + \log(\sqrt{x^2+1} - 1) \right] + C. \end{equation} Note that one can rewrite the log term as: \begin{equation} \log(\sqrt{x^2+1}-x) = - \log(\sqrt{x^2+1}+x) = - \sinh^{-1}(x). \end{equation}

  • 2
    $\begingroup$ Could you flesh out the steps when you substitute $y=\sqrt{u}-\sqrt{u-1}$ arriving in terms of $y^8$? $\endgroup$
    – user284001
    Aug 13, 2021 at 10:37
  • $\begingroup$ @SoboKevSpace I added some expressions needed for the variable transformation. $\endgroup$
    – tueda
    Aug 13, 2021 at 10:59
  • $\begingroup$ Can you please add the final result in terms of only $x$? $\endgroup$ Aug 14, 2021 at 21:26
  • $\begingroup$ @LouisCloete OK, I re-edited it and added the final result in $x$. $\endgroup$
    – tueda
    Aug 15, 2021 at 7:02

Proceeding with your method

$\begin{align} \int x^2\sqrt{x^2+1}\;dx &=\int \sqrt{x^2+1}\;x\cdot x\;dx\\ &=\frac1{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;\left( \text{ let $\begin{align} u&=x^2+1 \\ du&=2xdx\end{align}$}\right)\\ &=\frac1{2}\int \sqrt{u^2-u} \;du\\ &=\frac1{2} \int \sqrt{\left(u-\frac1{2}\right)^2-\left(\frac1{2}\right)^2} \;du \;\text{ (completing the square)} \\ &=\frac1{2}\left({\left(u-\frac1{2}\right)\over 2}{\sqrt{u^2-u}}-{\left(\frac1{2}\right)^2 \over 2}{ \ln \left|\left(u-\frac1{2}\right) +\sqrt{u^2-u}\right|} +C\right)\\ &\text{ (using $\int \sqrt{x^2-a^2} dx={x\over 2}{\sqrt{x^2-a^2}}-{a^2 \over 2}{ \ln |x+\sqrt{x^2-a^2}|} +C)$}\\ &={(2u-1)\over 8}{\sqrt{u^2-u}}-{\frac{ \ln |(2u-1) +2 \sqrt{u^2-u}|}{16}} +C'\\ &\text{ substituting $\left(\begin{align} u&=x^2+1 \\ 2u-1 &=2x^2 + 1 \\ \sqrt{u^2-u}&=\sqrt{(x^2 +1)^2-(x^2+1)}=\sqrt{x^4+x^2}\end{align}\right)$} \\ &=\frac{{(2x^2 + 1)}{\sqrt{x^4+x^2}}}{8}-{\frac{ \ln |(2x^2+1) +2 \sqrt{x^4+x^2}|}{16}} +C'\\ &=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}}{8}-{\frac{ \ln |(x^2+(x^2+1) +2x \sqrt{x^2+1}|}{16}} +C'\\ &=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}}{8}-{\frac{ \ln |(x+\sqrt{x^2+1})^2|}{16}} +C'\\ &=\frac{{(2x^3 + x)}{\sqrt{x^2+1}}-\ln |x+\sqrt{x^2+1}|}{8} +C' \end{align}$

  • $\begingroup$ @LouisCloete Yes, you might be missing something which is called "completing the square" to calculate the integral, considering that you stopped at $\int \sqrt{u}\;\sqrt{u-1}\;du$ which is of the form $\int \sqrt{au^2+bu+c} \;du$. Also, $\int \sqrt{x^2-a^2}\; dx$ is a standard integral which is well known and can be directly used where I read, but you might not be allowed to use it directly.You can refer to page 328 here ncert.nic.in/ncerts/l/lemh201.pdf for its proof. You can also try the proof using trigonometric identities yourself $\endgroup$ Aug 13, 2021 at 14:59
  • $\begingroup$ Trigonometric substitution was the magic phrase I needed. I am studying at a correspondence university while working full time, so I can't attend their online lectures for the most part. After I saw that term I found it in my textbook. I arrive at the same expression as you and @user773458 arrived, but by substituting $x=tan{\theta}$ $\endgroup$ Aug 14, 2021 at 21:23

As suggested in the comments we can use the substitution $x=\mbox{sinh}(u)$ and some other well known trig identities:

$1)$ $dx=(\mbox{sinh}(u))'du=\mbox{cosh}(u)du$

$2)$ $\mbox{cosh}^{2}(u)-\mbox{sinh}^{2}(u)=1$

$3)$ $\mbox{cosh}(u)=\frac{e^{u}+e^{-u}}{2}$

$4)$ $\mbox{arcsinh}(u)=\mbox{log}(u+\sqrt{1+u^{2}})$

$5)$ $\int \mbox{sinh}^{n}(u)du=\frac{\mbox{cosh}(u)\mbox{sinh}^{n-1}(u)}{n}-\frac{n-1}{n}\int \mbox{sinh}^{n-2}(u)du$

You can prove the last identity using integration by parts. So your integral becomes:

$\begin{align} \int x^{2}\sqrt{x^{2}+1}\;dx &=\int\mbox{sinh}^2(u)\sqrt{\mbox{sinh}^2(u)+1}\:\mbox{cosh}(u)\;du\\ &=\int \mbox{sinh}^{2}(u)\mbox{cosh}^{2}(u)\;du\\ &=\int \mbox{sinh}^{2}(u)(1+\mbox{sinh}^{2}(u))\;du\\ &=\int \mbox{sinh}^{2}(u)\;du\:+\:\int \mbox{sinh}^{4}(u)\;du\\ &=\int \mbox{sinh}^{2}(u)\;du\:+\:\begin{bmatrix}\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\;-\;\frac{3}{4}\int \mbox{sinh}^{2}(u)\;du\end{bmatrix}\\ &=\frac{1}{4}\int \mbox{sinh}^{2}(u)\;du\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\\ &=\frac{1}{4}\begin{bmatrix} \frac{\mbox{cosh}(u)\mbox{sinh}(u)}{2}-\frac{1}{2} \int du \end{bmatrix}\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\\ &=\bigstar \end{align}$

Let's compute something useful: $u=\mbox{arcsinh}(x)=\mbox{log}(x+\sqrt{1+x^{2}})$. So we have that: $\mbox{sinh}(u)=\mbox{sinh}(\mbox{arcsinh}(x))=x$

$\begin{align} \mbox{cosh}(u) &=\mbox{cosh}(\mbox{arcsinh}(x))\\ &=\frac{x+\sqrt{1+x^{2}}+\frac{1}{x+\sqrt{1+x^{2}}}}{2}\\ &=\frac{x^{2}+x\sqrt{1+x^{2}}+1}{x+\sqrt{1+x^{2}}}\\ &=\frac{x^{2}+x\sqrt{1+x^{2}}+1}{x+\sqrt{1+x^{2}}} \cdot \frac{\sqrt{1+x^{2}}-x}{\sqrt{1+x^{2}}-x}\\ &=...=\sqrt{1+x^{2}} \end{align}$

So the integral becomes:

$\begin{align} \bigstar &=\frac{\mbox{cosh}(u)\mbox{sinh}(u)-\mbox{arcsinh}(u)}{8}\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}+C\\ &=\frac{x\sqrt{1+x^{2}}\:-\:\mbox{log}(x+\sqrt{1+x^{2}})}{8}\:+\:\frac{x^{3}\sqrt{1+x^{2}}}{4}+C\\ &=\frac{(2x^{3}+x)\sqrt{1+x^{2}}\:-\:\mbox{log}(x+\sqrt{1+x^{2}})}{8}+C \end{align}$

  • 1
    $\begingroup$ I have unfortunately never been exposed to hyperbolic trig functions at school or (I think) in my Calculus course. (It's through Unisa, the world's first correspondence-only university, and I work full time, so I'm not sure what was maybe covered in the online lectures during my work hours). Both you and @Aman Kushwaha above arrived at the same answer I got, however. So I believe it is correct, we just skinned the cat in three different ways. $\endgroup$ Aug 14, 2021 at 21:19
  • $\begingroup$ Oh, I'm sorry you didn't study hyperbolic trig functions. Anyway... Thank you for upvoting my answer! Cheers :) $\endgroup$
    – user773458
    Aug 15, 2021 at 6:34
  • $\begingroup$ I did find hyperbolic trig functions in the textbook. Which probably means I need to find out what they are... Thanks for the help. $\endgroup$ Aug 15, 2021 at 11:45

I have used the trigonometric substitution $x=tan\theta$ and proceeded to get the same answer as @Aman Kushwaha got via completing the square and @user773458 got via hyperbolic trig identies. Below is my steps for reference:

$$ \int (x^2-\sqrt{x^2 +1})dx \\ =\int tan^2\theta \sqrt{tan^2 \theta + 1}sec^2 \theta \;d\theta \\ =\int tan^2 \theta \sqrt{sec^2 \theta}sec^2 \theta \;d\theta \\ =\int tan^2 \theta sec^3 \theta \;d\theta \\ =\int sec^3 \theta(sec^2 \theta - 1) \; d\theta \\ =\int sec^5 \theta - sec^3 \theta \; d\theta \\ =\int sec^5 \theta \; d\theta - \int sec^3 \theta \; d\theta \\ =sec^3 \theta \; tan \theta \; - \; 3 \int sec^3 \theta \; tan^2 \theta \; d\theta - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\ \therefore 4 \int sec^3 \theta \; tan^2 \theta \; d\theta = sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\ \therefore \int sec^3 \theta \; tan^2 \theta \; d\theta = \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta(sec^2 \theta - 1)\; d\theta\Big)\bigg) \\ = \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec^3 \theta \; d\theta \; - \int sec \theta \; d\theta\Big)\bigg) \\ = \frac{1}{4}\big(sec^3 \theta \; tan \theta \; - \; \frac{1}{2}(\sec \theta \; tan \theta - \ln|sec \theta + tan \theta|)\big) + C\\ =\frac{1}{4}\big(\sqrt{tan^2 \theta + 1}(tan^2 \theta + 1)tan \theta \; - \; \frac{1}{2}(\sqrt{tan^2 \theta + 1}(tan \theta) - \ln | \sqrt{tan^2 \theta + 1} + tan \theta|)\big) + C\\ =\frac{1}{4}\big(\sqrt{x^2 + 1}(x^2 + 1)x \; - \; \frac{1}{2}(\sqrt{x^2 + 1}(x) - \ln | \sqrt{x^2 + 1} + x|)\big) + C\\ =\frac{(2x^3+2x)\sqrt{x^2 +1}-x\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\ =\frac{(2x^3 + x)\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\ $$


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