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I'm new of the site. I must solve this exercise:

$$\int \frac{1}{x^2 \sqrt{x^2+9}}\,dx$$

I tried every substitution, but I didn't reach that I want. Can you help me, please?

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  • $\begingroup$ Did you take $x = 3 \cosh \alpha$? $\endgroup$ – Mattos Feb 12 '15 at 11:48
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    $\begingroup$ Please be specific about what you have tried. You cannot have tried every posible substitution, for there are infinitely many! :) $\endgroup$ – Shaun Feb 12 '15 at 11:50
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Here is an approach.

Assuming that $x\geq b>0. $ You may write $$ \begin{align} \int \frac{1}{x^2 \sqrt{x^2+9}}dx &=\int \frac{1}{x^2 \sqrt{1+\frac{9}{x^2}}}\frac{dx}{x}\\\\ &=-\int \frac{u}{\sqrt{1+9u^2}}\:du\\\\ &=-\frac19\sqrt{1+9u^2}\\\\ &=-\frac19\frac{\sqrt{x^2+9}}{x}, \end{align} $$ the case $x\leq a<0 $ is similar.

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  • $\begingroup$ On the RHS of your first equation, that should be an $x^3$ on the outside of the square root, not $x^2$. See Mike's solution. $\endgroup$ – Paddling Ghost Feb 24 '15 at 16:47
  • $\begingroup$ @user3440448 Please, look carefully you will see that you have $x^2$ in front of the radical but also $x$ back to it. Thanks. $\endgroup$ – Olivier Oloa Feb 24 '15 at 16:49
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    $\begingroup$ Yes I see it now. I'm gonna leave this comments here in case anyone else missed it as I did. Thanks. $\endgroup$ – Paddling Ghost Feb 24 '15 at 16:51
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Let $x = 3\tan(u).$ Therefore $\frac{dx}{du} = 3\sec^2{u} \implies dx=3\sec^2{u}\ du$. So, our integral becomes

$$\int\frac{3\sec^2{u}}{9\tan^2(u)\sqrt{9\tan^2(u)+9}}du= \int\frac{\sec^2{u}}{9\tan^2(u)\sqrt{\tan^2(u)+1}}du$$

$$=\int\frac{\sec{u}}{9\tan^2(u)}du$$ $$=\frac{1}{9}\int\frac{\sec{u}\cos^{2}(u)}{\sin^2(u)}du$$ $$=\frac{1}{9}\int\frac{\cos(u)}{\sin^2(u)}du.$$ $$=\frac{1}{9}\int\frac{1}{\sin^2(u)}d(\sin{u})$$ $$=\frac{1}{9}\int\sin^{-2}(u)d(\sin{u})$$ $$=\frac{1}{9}\left[-\sin^{-1}(u)\right]+C$$

Now we must return our expression in terms of $x$. That is, we wish to find the value of $\sin^{-1}(u)$ in terms of $x$. Note that this is not sine inverse - it is "sin(u) to the power of -1". Recall that:

$$ x = 3\tan{u} \implies \tan{u} = \frac{x}{3}.$$

Assume that $0 < u < \frac{\pi}{2}$ if you draw a right angled triangle you will notice that the opposite side is $x$ and the adjacent side is $3$. By Pythagoras's Theorem the hypotenuse is therefore $\sqrt{x^{2}+9}$. Therefore

$$\sin{u} = \frac{x}{\sqrt{x^{2}+9}}.$$

Thus, our answer is

$$\frac{1}{9}\left[-\frac{x}{\sqrt{x^{2}+9}}\right]^{-1}+C = -\frac{\sqrt{x^{2}+9}}{9x} +C .$$

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  • $\begingroup$ Thanks, but now, calculating, I have $\frac{1}{9 \operatorname{sin} \operatorname{arctg} \frac{x}{3}}$. The solution in the book is $-\frac{\sqrt{x^2+9}}{9x}$. $\endgroup$ – user214941 Feb 12 '15 at 12:01
  • $\begingroup$ @userNew. Ok. I will edit my post and complete the solution for you. $\endgroup$ – Gustavo Louis G. Montańo Feb 12 '15 at 12:02
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Either one of the substitutions

should work.

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I think Olivier had the right idea, but his substitution went awry. Similar to how he started.

$$\int\frac{dx}{x^2\sqrt{x^2+9}}=\int\frac{dx}{x^3\sqrt{1+9x^{-2}}}$$

Now substitute

$$u=1+9x^{-2},du=-18x^{-3}dx$$

$$-\frac1{18}\int u^{-1/2}du=-\frac19u^{1/2}+C=-\frac19\sqrt{1+9x^{-2}}+C$$

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Hint: Convert to $1\over x^2\sqrt{x^2+1}$ form first.

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