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I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$

I said that $u(x) = x^2 + 1$

and then that $dx = 2x\,dx$ so I rewrote the integral as

$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$

which is also

$$\int2x^4\sqrt{x^2 +1} \,dx$$

and then it is easy to integrate, is that all legal to do?

Michael Hardy
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Jacob C
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  • It appears that you started to try a $u$ substitution but then switched to a different plan. What happens if you complete the $u$ substitution? Do you have a text you are studying? – abiessu Sep 17 '13 at 02:34
  • Well I do i have a textbook for my class but this problem is simply one of many practice problems listed in the section we are working on.There's nothing about this problem in the book that i can see... I'm deep into the practice problems were it requires applied knowledge and such, I'm not sure what you mean. – Jacob C Sep 17 '13 at 02:39
  • Try the Maple command IntTutor which gives hints and shows the required solution step by step. – user64494 Sep 20 '13 at 16:53

4 Answers4

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I'm afraid not. You should not have an equation with a "differential factor" on one side and not on the other (that is, $dx=2x$ is nonsense). For more on how to deal with differential factors, you might find the second part of this answer (from "Now, if I wrote..." through "...let's get back to your problem.") useful.

What you can say is that $$\frac{du}{dx}=2x,$$ so that $$du=2x\,dx,$$ so that $$x\,dx=\frac12\,du.$$ Then your substitution gives you $$\int x^3\sqrt{x^2+1}\,dx=\frac12\int x^2\sqrt{u}\,du.$$ We're not quite there, yet, though. Can you rewrite $x^2$ in terms of $u$?

Community
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Cameron Buie
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  • yeah, (u-1) then say (u^1.5 - u^.5) thats simple. thanks i need to remember to think things through and actually do the equation for derivatives since i seem to not understand it perfectly – Jacob C Sep 17 '13 at 02:49
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    That's it, yes! By the way, welcome to Math.SE! Just a few quick notes for you. For help formatting your questions (the better the question is formatted, the better the response is, usually), see here and follow the links to the guide. Kudos for putting your thoughts and attempts into your post (that is strongly encouraged)! – Cameron Buie Sep 17 '13 at 02:58
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\begin{align} &\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x = \int x^{2}\,{\rm d}\left[{1 \over 3}\,\left(x^{2} + 1\right)^{3/2}\right] \\[3mm]&= x^{2}\,{1 \over 3}\left(x^{2} + 1\right)^{3/2} - \int{1 \over 3}\left(x^{2} + 1\right)^{3/2} \,{\rm d}\left(x^{2} + 1\right) \\[3mm]&= {1 \over 3}\,x^{2}\left(x^{2} + 1\right)^{3/2} - {1 \over 3}\,{\left(x^{2} + 1\right)^{5/2} \over 5/2} = \left(x^{2} + 1\right)^{3/2}\left[% {1 \over 3}\,x^{2} - {2 \over 15}\left(x^{2} + 1\right) \right] \\[3mm]&= \left(x^{2} + 1\right)^{3/2}\,{3x^{2} - 2 \over 15} \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x \color{#000000}{\ =\ } {1 \over 15}\left(3x^{2} - 2\right)\left(x^{2} + 1\right)^{3/2}\ +\ \color{#000000}{\mbox{constant}} \quad} \\ \\ \hline \end{array} $$

Felix Marin
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HINT: when completing a $u$ substitution, the typical flow is as follows:

$$\int x^3\sqrt{x^2+1}dx$$

Substitute $u=x^2+1$, then $du=2xdx$ and we have $x^2=u-1$:

$$\int {(u-1)\sqrt u du\over 2}$$

From there, the remaining steps are to integrate, then reverse-substitute to obtain the function relative to $x$, and to add the unknown constant.

abiessu
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  • I wish I knew why this answer was down-voted. I think I have correctly shown the steps up to just before the point of potentially solving the problem for the OP. – abiessu Sep 21 '13 at 22:10
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Your substitution should be completed by letting $$\frac12\int (u-1) \sqrt{u}du$$

By using integration by parts, this can be solved.

J.H.
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